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Номер 838, страница 212 - гдз по геометрии 10-11 класс учебник Атанасян, Бутузов
Авторы: Атанасян Л. С., Бутузов В. Ф., Кадомцев С. Б., Позняк Э. Г., Киселёва Л. С.
Тип: Учебник
Издательство: Просвещение
Год издания: 2019 - 2026
Уровень обучения: базовый и углублённый
Цвет обложки: коричневый с ромбами
ISBN: 978-5-09-103606-0 (2023)
Допущено Министерством просвещения Российской Федерации
Математика: алгебра и начала математического анализа, геометрия
Популярные ГДЗ в 10 классе
Глава 8. Некоторые сведения из планиметрии. Параграф 2. Решение треугольников - номер 838, страница 212.
App\Models\Task {#1030 // resources/views/models/task/default.blade.php #connection: "mysql" #table: "tasks" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:24 [ "id" => 745253 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_page_start" => "212" "field_page_end" => null "field_url" => "/10-klass/geometrija/atanasjan-uchebnik/00-838" "field_display_title" => "838" "field_outside_task" => null "field_task_type" => Illuminate\Database\Eloquent\Collection {#1043 #items: array:1 [ 0 => App\Models\Term {#1042 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 26 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "номер" "field_cases" => array:6 [ "field_accusative_case" => "номер" "field_creative_case" => "номером" "field_dative_case" => "номеру" "field_genitive_case" => "номера" "field_nominative_case" => "номер" "field_prepositional_case" => "номере" ] "field_short_name" => "№" ] #original: array:6 [ "id" => 26 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "номер" "field_cases" => array:6 [ "field_accusative_case" => "номер" "field_creative_case" => "номером" "field_dative_case" => "номеру" "field_genitive_case" => "номера" "field_nominative_case" => "номер" "field_prepositional_case" => "номере" ] "field_short_name" => "№" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_match" => null "breadcrumbs" => [] "edition_groups" => Illuminate\Database\Eloquent\Collection {#1034 #items: [] #escapeWhenCastingToString: false } "top_parent_branch" => Illuminate\Database\Eloquent\Collection {#1047 #items: array:1 [ 0 => App\Models\Branch {#1046 #connection: "mysql" #table: "branches" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:24 [ "id" => 744411 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Некоторые сведения из планиметрии" "field_branch_order" => "8" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1051 #items: array:1 [ 0 => App\Models\Term {#1048 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:7 [ "id" => 29 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Глава" "field_cases" => array:6 [ "field_accusative_case" => "Главу" "field_creative_case" => "Главой" "field_dative_case" => "Главе" "field_genitive_case" => "Главы" "field_nominative_case" => "Глава" "field_prepositional_case" => "Главе" ] "field_page_check_not_needed" => null "field_short_name" => null ] #original: array:7 [ "id" => 29 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Глава" "field_cases" => array:6 [ "field_accusative_case" => "Главу" "field_creative_case" => "Главой" "field_dative_case" => "Главе" "field_genitive_case" => "Главы" "field_nominative_case" => "Глава" "field_prepositional_case" => "Главе" ] "field_page_check_not_needed" => null "field_short_name" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_page_start" => "194" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1087 #items: array:1 [ 0 => App\Models\Book {#1050 #connection: "mysql" #table: "books" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:50 [ "id" => 36 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_tree_default_status" => "tasks" "field_pages_status" => null "field_subject" => Illuminate\Database\Eloquent\Collection {#1052 #items: array:1 [ 0 => App\Models\Term {#1049 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:10 [ "id" => 6477 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Геометрия" "field_abbreviated_name" => null "field_cases" => array:6 [ …6] "field_foreign_lang_name" => null "field_short_name" => null "field_subject_type" => "technical_subject" "field_translit" => "geometrija" ] #original: array:10 [ "id" => 6477 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Геометрия" "field_abbreviated_name" => null "field_cases" => array:6 [ …6] "field_foreign_lang_name" => null "field_short_name" => null "field_subject_type" => "technical_subject" "field_translit" => "geometrija" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_class" => Illuminate\Database\Eloquent\Collection {#1053 #items: array:2 [ 0 => App\Models\Term {#1054 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 5459 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "10" "field_cases" => array:6 [ …6] "field_translit" => "desjatyj" ] #original: array:6 [ "id" => 5459 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "10" "field_cases" => array:6 [ …6] "field_translit" => "desjatyj" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 1 => App\Models\Term {#1055 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 5460 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "11" "field_cases" => array:6 [ …6] "field_translit" => "odinadcatyj" ] #original: array:6 [ "id" => 5460 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "11" "field_cases" => array:6 [ …6] "field_translit" => "odinadcatyj" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_publisher" => Illuminate\Database\Eloquent\Collection {#1056 #items: array:1 [ 0 => App\Models\Term {#1057 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 5153 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Просвещение" "field_cases" => null "field_translit" => "prosveschenie" ] #original: array:6 [ "id" => 5153 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Просвещение" "field_cases" => null "field_translit" => "prosveschenie" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_author" => Illuminate\Database\Eloquent\Collection {#1058 #items: array:5 [ 0 => App\Models\Term {#1059 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:12 [ "id" => 3753 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Атанасян" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Левон" "field_patronymic" => "Сергеевич" "field_surname_rp" => null ] #original: array:12 [ "id" => 3753 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Атанасян" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Левон" "field_patronymic" => "Сергеевич" "field_surname_rp" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 1 => App\Models\Term {#1060 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:12 [ "id" => 3949 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Бутузов" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Валентин" "field_patronymic" => "Фёдорович" "field_surname_rp" => null ] #original: array:12 [ "id" => 3949 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Бутузов" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Валентин" "field_patronymic" => "Фёдорович" "field_surname_rp" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 2 => App\Models\Term {#1061 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:12 [ "id" => 4518 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Кадомцев" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Сергей" "field_patronymic" => "Борисович" "field_surname_rp" => null ] #original: array:12 [ "id" => 4518 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Кадомцев" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Сергей" "field_patronymic" => "Борисович" "field_surname_rp" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 3 => App\Models\Term {#1062 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:12 [ "id" => 5494 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Позняк" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Эдуард" "field_patronymic" => "Генрихович" "field_surname_rp" => null ] #original: array:12 [ "id" => 5494 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Позняк" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Эдуард" "field_patronymic" => "Генрихович" "field_surname_rp" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 4 => App\Models\Term {#1063 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:12 [ "id" => 6905 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Киселёва" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Людмила" "field_patronymic" => "Сергеевна" "field_surname_rp" => null ] #original: array:12 [ "id" => 6905 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Киселёва" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Людмила" "field_patronymic" => "Сергеевна" "field_surname_rp" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_author_foreign" => Illuminate\Database\Eloquent\Collection {#1064 #items: [] #escapeWhenCastingToString: false } "field_book_type" => Illuminate\Database\Eloquent\Collection {#1065 #items: array:1 [ 0 => App\Models\Term {#1066 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:10 [ "id" => 6671 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Учебник" "field_book_type_foreign" => null "field_cases" => array:6 [ …6] "field_plural_form" => null "field_short_name" => null "field_short_name_foreign" => null "field_translit" => "uchebnik" ] #original: array:10 [ "id" => 6671 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Учебник" "field_book_type_foreign" => null "field_cases" => array:6 [ …6] "field_plural_form" => null "field_short_name" => null "field_short_name_foreign" => null "field_translit" => "uchebnik" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_country" => Illuminate\Database\Eloquent\Collection {#1067 #items: array:1 [ 0 => App\Models\Term {#1068 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 9 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Россия" "field_cases" => array:6 [ …6] "field_translit" => "rossiya" ] #original: array:6 [ "id" => 9 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Россия" "field_cases" => array:6 [ …6] "field_translit" => "rossiya" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_city" => Illuminate\Database\Eloquent\Collection {#1069 #items: array:1 [ 0 => App\Models\Term {#1070 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 15 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Москва" "field_cases" => array:6 [ …6] "field_translit" => "moskva" ] #original: array:6 [ "id" => 15 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Москва" "field_cases" => array:6 [ …6] "field_translit" => "moskva" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_series" => Illuminate\Database\Eloquent\Collection {#1071 #items: [] #escapeWhenCastingToString: false } "field_umk" => Illuminate\Database\Eloquent\Collection {#1072 #items: [] #escapeWhenCastingToString: false } "field_level_of_education" => Illuminate\Database\Eloquent\Collection {#1073 #items: array:1 [ 0 => App\Models\Term {#1074 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 41 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "базовый и углублённый" "field_cases" => array:6 [ …6] "field_translit" => "bazovyy i uglublennyy" ] #original: array:6 [ "id" => 41 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "базовый и углублённый" "field_cases" => array:6 [ …6] "field_translit" => "bazovyy i uglublennyy" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_standart_of_education" => Illuminate\Database\Eloquent\Collection {#1075 #items: array:1 [ 0 => App\Models\Term {#1076 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:10 [ "id" => 6706 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "ФГОС (старый)" "field_cases" => array:6 [ …6] "field_color" => null "field_end_year" => "2022" "field_start_year" => "2019" "field_translit" => "fgos-old" "field_type" => null ] #original: array:10 [ "id" => 6706 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "ФГОС (старый)" "field_cases" => array:6 [ …6] "field_color" => null "field_end_year" => "2022" "field_start_year" => "2019" "field_translit" => "fgos-old" "field_type" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_publication_number" => "11" "field_publication_type" => Illuminate\Database\Eloquent\Collection {#1077 #items: array:1 [ 0 => App\Models\Term {#1078 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 33 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "стереотипное" "field_cases" => array:6 [ …6] "field_translit" => "stereotipnoe" ] #original: array:6 [ "id" => 33 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "стереотипное" "field_cases" => array:6 [ …6] "field_translit" => "stereotipnoe" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_under_the_edition" => Illuminate\Database\Eloquent\Collection {#1079 #items: [] #escapeWhenCastingToString: false } "field_under_the_edition_degree" => null "field_cover_description" => "с ромбами" "field_publication_year" => "2019" "field_publication_year_until" => null "field_part" => "" "field_part_writing" => "" "field_second_foreign_language" => null "field_for_whom" => "Учебник для общеобразовательных организаций" "field_allowed" => "Допущено Министерством просвещения Российской Федерации" "field_reserve_field" => "Математика: алгебра и начала математического анализа, геометрия" "field_link_to_source" => null "field_tasks_count" => "1147" "field_priority" => "12" "field_default_folder" => "/geometrija_10/atanasjan-u23/" "field_isbn" => "978-5-09-103606-0 (2023)" "field_cover" => array:1 [ 0 => "/media/geometrija_10/atanasjan-u23/covers/cover1.webp?ts=1739363242" ] "field_cover_alts" => array:1 [ 0 => "" ] "field_covers" => array:1 [ 0 => array:4 [ "path" => "/media/geometrija_10/atanasjan-u23/covers/cover1.webp?ts=1739363242" "alt" => "" "width" => "1200" "height" => "1661" ] ] "field_popular_book" => null "field_recommended_books" => Illuminate\Database\Eloquent\Collection {#1080 #items: [] #escapeWhenCastingToString: false } "field_new_book" => Illuminate\Database\Eloquent\Collection {#1081 #items: [] #escapeWhenCastingToString: false } "field_old_book" => Illuminate\Database\Eloquent\Collection {#1082 #items: [] #escapeWhenCastingToString: false } "field_url" => "/10-klass/geometrija/atanasjan-uchebnik" "field_cover_color" => Illuminate\Database\Eloquent\Collection {#1083 #items: array:1 [ 0 => App\Models\Term {#1084 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 51 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "коричневый" "field_cases" => array:6 [ …6] "field_translit" => "korichnevyy" ] #original: array:6 [ "id" => 51 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "коричневый" "field_cases" => array:6 [ …6] "field_translit" => "korichnevyy" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "breadcrumbs" => array:3 [ "class" => 381 "subject" => 543 "class_subject" => 392 ] ] #original: array:50 [ "id" => 36 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_tree_default_status" => "tasks" "field_pages_status" => null "field_subject" => Illuminate\Database\Eloquent\Collection {#1052} "field_class" => Illuminate\Database\Eloquent\Collection {#1053} "field_publisher" => Illuminate\Database\Eloquent\Collection {#1056} "field_author" => Illuminate\Database\Eloquent\Collection {#1058} "field_author_foreign" => Illuminate\Database\Eloquent\Collection {#1064} "field_book_type" => Illuminate\Database\Eloquent\Collection {#1065} "field_country" => Illuminate\Database\Eloquent\Collection {#1067} "field_city" => Illuminate\Database\Eloquent\Collection {#1069} "field_series" => Illuminate\Database\Eloquent\Collection {#1071} "field_umk" => Illuminate\Database\Eloquent\Collection {#1072} "field_level_of_education" => Illuminate\Database\Eloquent\Collection {#1073} "field_standart_of_education" => Illuminate\Database\Eloquent\Collection {#1075} "field_publication_number" => "11" "field_publication_type" => Illuminate\Database\Eloquent\Collection {#1077} "field_under_the_edition" => Illuminate\Database\Eloquent\Collection {#1079} "field_under_the_edition_degree" => null "field_cover_description" => "с ромбами" "field_publication_year" => "2019" "field_publication_year_until" => null "field_part" => "" "field_part_writing" => "" "field_second_foreign_language" => null "field_for_whom" => "Учебник для общеобразовательных организаций" "field_allowed" => "Допущено Министерством просвещения Российской Федерации" "field_reserve_field" => "Математика: алгебра и начала математического анализа, геометрия" "field_link_to_source" => null "field_tasks_count" => "1147" "field_priority" => "12" "field_default_folder" => "/geometrija_10/atanasjan-u23/" "field_isbn" => "978-5-09-103606-0 (2023)" "field_cover" => array:1 [ 0 => "/media/geometrija_10/atanasjan-u23/covers/cover1.webp?ts=1739363242" ] "field_cover_alts" => array:1 [ 0 => "" ] "field_covers" => array:1 [ 0 => array:4 [ "path" => "/media/geometrija_10/atanasjan-u23/covers/cover1.webp?ts=1739363242" "alt" => "" "width" => "1200" "height" => "1661" ] ] "field_popular_book" => null "field_recommended_books" => Illuminate\Database\Eloquent\Collection {#1080} "field_new_book" => Illuminate\Database\Eloquent\Collection {#1081} "field_old_book" => Illuminate\Database\Eloquent\Collection {#1082} "field_url" => "/10-klass/geometrija/atanasjan-uchebnik" "field_cover_color" => Illuminate\Database\Eloquent\Collection {#1083} "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "breadcrumbs" => array:3 [ "class" => 381 "subject" => 543 "class_subject" => 392 ] ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "branch_parent" => Illuminate\Database\Eloquent\Collection {#1085 #items: [] #escapeWhenCastingToString: false } ] #original: array:24 [ "id" => 744411 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Некоторые сведения из планиметрии" "field_branch_order" => "8" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1051} "field_page_start" => "194" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1087} "branch_parent" => Illuminate\Database\Eloquent\Collection {#1085} ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "parent_branches" => Illuminate\Database\Eloquent\Collection {#1122 #items: array:2 [ 0 => App\Models\Branch {#1130 #connection: "mysql" #table: "branches" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:24 [ "id" => 744411 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Некоторые сведения из планиметрии" "field_branch_order" => "8" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1131 #items: array:1 [ 0 => App\Models\Term {#1048} ] #escapeWhenCastingToString: false } "field_page_start" => "194" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1132 #items: array:1 [ 0 => App\Models\Book {#1050} ] #escapeWhenCastingToString: false } "branch_parent" => Illuminate\Database\Eloquent\Collection {#1133 #items: [] #escapeWhenCastingToString: false } ] #original: array:24 [ "id" => 744411 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Некоторые сведения из планиметрии" "field_branch_order" => "8" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1131} "field_page_start" => "194" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1132} "branch_parent" => Illuminate\Database\Eloquent\Collection {#1133} ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 1 => App\Models\Branch {#1134 #connection: "mysql" #table: "branches" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:24 [ "id" => 744413 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Решение треугольников" "field_branch_order" => "2" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1138 #items: array:1 [ 0 => App\Models\Term {#1135 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:7 [ "id" => 32 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Параграф" "field_cases" => array:6 [ "field_accusative_case" => "Параграф" …5 ] "field_page_check_not_needed" => null "field_short_name" => "§" ] #original: array:7 [ "id" => 32 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Параграф" "field_cases" => array:6 [ …6] "field_page_check_not_needed" => null "field_short_name" => "§" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_page_start" => "202" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1137 #items: array:1 [ 0 => App\Models\Book {#1050} ] #escapeWhenCastingToString: false } "branch_parent" => Illuminate\Database\Eloquent\Collection {#1136 #items: array:1 [ 0 => App\Models\Branch {#1046} ] #escapeWhenCastingToString: false } ] #original: array:24 [ "id" => 744413 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Решение треугольников" "field_branch_order" => "2" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1138} "field_page_start" => "202" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1137} "branch_parent" => Illuminate\Database\Eloquent\Collection {#1136} ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "content" => Illuminate\Database\Eloquent\Collection {#1036 #items: array:4 [ 0 => App\Models\Element {#1112 #connection: "mysql" #table: "elements" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:7 [ "id" => 940166 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1101 #items: array:1 [ 0 => App\Models\Edition {#1109 #connection: "mysql" #table: "editions" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 2315 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Условие" "field_order" => "1" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1107 …2} "field_content_type" => "free" "field_content_mode" => "text, image" "field_page_content_mode" => "image" "field_content_text_checked" => null "field_page_content_text_checked" => "0" "field_solution_author" => "Автор" "field_moderator" => "pasha" "field_edition_type" => "statement" "field_root_dir" => "0-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1108 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1106 …2} "field_process_formula" => "wiris" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1105 …2} "field_content_source" => null ] #original: array:21 [ "id" => 2315 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Условие" "field_order" => "1" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1107 …2} "field_content_type" => "free" "field_content_mode" => "text, image" "field_page_content_mode" => "image" "field_content_text_checked" => null "field_page_content_text_checked" => "0" "field_solution_author" => "Автор" "field_moderator" => "pasha" "field_edition_type" => "statement" "field_root_dir" => "0-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1108 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1106 …2} "field_process_formula" => "wiris" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1105 …2} "field_content_source" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "task" => array:2 [ "refs" => "745253" "type" => "task" ] "text" => """ <p><strong>838.</strong> Биссектрисы <em>АА₁, ВВ₁</em> и <em>СС₁</em> треугольника <em>АВС</em> со сторонами <span class="long"><em>АВ = с,</em></span> <span class="long"><em>ВС = а</em></span> и <span class="long"><em>СА = b</em></span> пересекаются в точке <em>О</em>.</p><figure><figure>\n <img src="/media/geometrija_10/atanasjan-u23/0-00/838.webp?ts=1745834741" alt="Доказать, что одна из биссектрис делится точкой О" loading="lazy" width="1409" height="333">\n </figure>\n <figcaption> </figcaption></figure><p>в) Может ли хотя бы одна из биссектрис треугольника делиться точкой О пополам? г) Докажите, что одна из биссектрис делится точкой О в отношении <span class="long">2 : 1,</span> считая от вершины, тогда и только тогда, когда одна из сторон треугольника равна полусумме двух других сторон.</p> """ "img" => array:1 [ 0 => array:5 [ "name" => "838-1.jpg" "alt" => null "width" => "1549" "height" => 531 "path" => "/media/geometrija_10/atanasjan-u23/0-00/838-1.webp?ts=1739436936" ] ] ] #original: array:7 [ "id" => 940166 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1101} "task" => array:2 [ "refs" => "745253" "type" => "task" ] "text" => """ <p><strong>838.</strong> Биссектрисы <em>АА₁, ВВ₁</em> и <em>СС₁</em> треугольника <em>АВС</em> со сторонами <span class="long"><em>АВ = с,</em></span> <span class="long"><em>ВС = а</em></span> и <span class="long"><em>СА = b</em></span> пересекаются в точке <em>О</em>.</p><figure><figure>\n <img src="/media/geometrija_10/atanasjan-u23/0-00/838.webp?ts=1745834741" alt="Доказать, что одна из биссектрис делится точкой О" loading="lazy" width="1409" height="333">\n </figure>\n <figcaption> </figcaption></figure><p>в) Может ли хотя бы одна из биссектрис треугольника делиться точкой О пополам? г) Докажите, что одна из биссектрис делится точкой О в отношении <span class="long">2 : 1,</span> считая от вершины, тогда и только тогда, когда одна из сторон треугольника равна полусумме двух других сторон.</p> """ "img" => array:1 [ 0 => array:5 [ "name" => "838-1.jpg" "alt" => null "width" => "1549" "height" => 531 "path" => "/media/geometrija_10/atanasjan-u23/0-00/838-1.webp?ts=1739436936" ] ] ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 1 => App\Models\Element {#1102 #connection: "mysql" #table: "elements" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 942799 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1089 #items: array:1 [ 0 => App\Models\Edition {#1099 #connection: "mysql" #table: "editions" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 2317 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 2" "field_order" => "3" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1092 …2} "field_content_type" => "free" "field_content_mode" => "image" 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array:2 [ "refs" => "745253" "type" => "task" ] "text" => "<p><strong>а)</strong></p><p>Рассмотрим треугольник $ABC$ со сторонами $AB=c$, $BC=a$, $CA=b$. Биссектрисы $AA_1, BB_1, CC_1$ пересекаются в точке $O$.</p><p>Чтобы найти отношение $\frac{AO}{OA_1}$, рассмотрим треугольник $AB B_1$. В этом треугольнике $AO$ является биссектрисой угла $A$. По свойству биссектрисы, она делит противолежащую сторону $BB_1$ в отношении, равном отношению прилежащих сторон:</p><p>$\frac{AO}{OB_1} = \frac{AB}{AB_1}$. Ой, это не то отношение.</p><p>Воспользуемся другим подходом. Рассмотрим треугольник $ACA_1$. В нем $CO$ является биссектрисой угла $C$. По свойству биссектрисы, она делит противолежащую сторону $AA_1$ в отношении:</p><p>$\frac{AO}{OA_1} = \frac{AC}{CA_1}$</p><p>Теперь найдем длину отрезка $CA_1$. $AA_1$ – биссектриса угла $A$ в треугольнике $ABC$. По свойству биссектрисы треугольника:</p><p>$\frac{BA_1}{CA_1} = \frac{AB}{AC} = \frac{c}{b}$</p><p>Мы знаем, что $BA_1 + CA_1 = BC = a$. Из пропорции имеем $BA_1 = \frac{c}{b} CA_1$. Подставим в сумму:</p><p>$\frac{c}{b} CA_1 + CA_1 = a$</p><p>$CA_1 \left(\frac{c}{b} + 1\right) = a$</p><p>$CA_1 \left(\frac{c+b}{b}\right) = a$</p><p>$CA_1 = \frac{ab}{b+c}$</p><p>Теперь подставим $CA_1$ в формулу для искомого отношения:</p><p>$\frac{AO}{OA_1} = \frac{AC}{CA_1} = \frac{b}{\frac{ab}{b+c}} = \frac{b(b+c)}{ab} = \frac{b+c}{a}$</p><p>Аналогично, применяя свойство биссектрисы к другим треугольникам, можно получить остальные отношения.Рассмотрим треугольник $BCB_1$, в нем $AO$ — биссектриса угла $A$. Нет, это неверно.</p><p>Рассмотрим треугольник $ABB_1$. $AO$ является биссектрисой угла $A$. Тогда $\frac{BO}{OB_1} = \frac{AB}{AB_1}$. Найдем $AB_1$. По свойству биссектрисы $BB_1$ в треугольнике $ABC$: $\frac{AB_1}{CB_1} = \frac{AB}{BC} = \frac{c}{a}$. $AB_1 + CB_1 = AC = b$. Отсюда $CB_1 = b - AB_1$.$\frac{AB_1}{b-AB_1} = \frac{c}{a} \Rightarrow a \cdot AB_1 = c(b-AB_1) \Rightarrow (a+c)AB_1 = bc \Rightarrow AB_1 = \frac{bc}{a+c}$.Тогда $\frac{BO}{OB_1} = \frac{c}{\frac{bc}{a+c}} = \frac{a+c}{b}$.</p><p>Рассмотрим треугольник $ACC_1$. $BO$ не является биссектрисой. Рассмотрим треугольник $BCC_1$. $BO$ - биссектриса угла $B$. Тогда $\frac{CO}{OC_1} = \frac{BC}{BC_1}$. Найдем $BC_1$. По свойству биссектрисы $CC_1$ в треугольнике $ABC$: $\frac{AC_1}{BC_1} = \frac{AC}{BC} = \frac{b}{a}$. $AC_1 + BC_1 = AB = c$. Отсюда $AC_1 = c - BC_1$.$\frac{c-BC_1}{BC_1} = \frac{b}{a} \Rightarrow a(c-BC_1) = b \cdot BC_1 \Rightarrow ac = (a+b)BC_1 \Rightarrow BC_1 = \frac{ac}{a+b}$.Тогда $\frac{CO}{OC_1} = \frac{a}{\frac{ac}{a+b}} = \frac{a+b}{c}$.</p><p>Ответ: $\frac{AO}{OA_1} = \frac{b+c}{a}$, $\frac{BO}{OB_1} = \frac{a+c}{b}$, $\frac{CO}{OC_1} = \frac{a+b}{c}$.</p><p><strong>б)</strong></p><p><strong>1. Докажем, что $\frac{AO}{AA_1} + \frac{BO}{BB_1} + \frac{CO}{CC_1} = 2$.</strong></p><p>Используем отношения, найденные в пункте <strong>а)</strong>.Из $\frac{AO}{OA_1} = \frac{b+c}{a}$ следует, что $OA_1 = AO \cdot \frac{a}{b+c}$.Поскольку $AA_1 = AO + OA_1$, то $AA_1 = AO + AO \cdot \frac{a}{b+c} = AO \left(1 + \frac{a}{b+c}\right) = AO \left(\frac{b+c+a}{b+c}\right)$.Отсюда получаем отношение $\frac{AO}{AA_1} = \frac{b+c}{a+b+c}$.</p><p>Аналогично для двух других биссектрис:$\frac{BO}{BB_1} = \frac{a+c}{a+b+c}$</p><p>$\frac{CO}{CC_1} = \frac{a+b}{a+b+c}$</p><p>Сложим эти три отношения:</p><p>$\frac{AO}{AA_1} + \frac{BO}{BB_1} + \frac{CO}{CC_1} = \frac{b+c}{a+b+c} + \frac{a+c}{a+b+c} + \frac{a+b}{a+b+c} = \frac{(b+c) + (a+c) + (a+b)}{a+b+c} = \frac{2a+2b+2c}{a+b+c} = \frac{2(a+b+c)}{a+b+c} = 2$.</p><p>Что и требовалось доказать.</p><p><strong>2. Докажем, что $\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1} = 1$.</strong>(Предполагается, что в условии опечатка и имеется в виду $OA_1$, а не $AO_1$).</p><p>Это можно доказать двумя способами.</p><p>Способ 1. Выразим отношение $\frac{OA_1}{AA_1}$.Так как $AA_1 = AO + OA_1$, то $\frac{OA_1}{AA_1} = \frac{OA_1}{AO+OA_1}$. Разделим числитель и знаменатель на $OA_1$:</p><p>$\frac{OA_1}{AA_1} = \frac{1}{\frac{AO}{OA_1} + 1}$.</p><p>Используя результат из пункта <strong>а)</strong> $\frac{AO}{OA_1} = \frac{b+c}{a}$:</p><p>$\frac{OA_1}{AA_1} = \frac{1}{\frac{b+c}{a} + 1} = \frac{1}{\frac{b+c+a}{a}} = \frac{a}{a+b+c}$.</p><p>Аналогично: $\frac{OB_1}{BB_1} = \frac{b}{a+b+c}$ и $\frac{OC_1}{CC_1} = \frac{c}{a+b+c}$.</p><p>Сложим эти три отношения:</p><p>$\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1} = \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c} = \frac{a+b+c}{a+b+c} = 1$.</p><p>Что и требовалось доказать.</p><p>Способ 2. Используем уже доказанное тождество.</p><p>$\frac{AO}{AA_1} + \frac{BO}{BB_1} + \frac{CO}{CC_1} = 2$.</p><p>Заменим $AO = AA_1 - OA_1$, $BO = BB_1 - OB_1$, $CO = CC_1 - OC_1$:</p><p>$\frac{AA_1 - OA_1}{AA_1} + \frac{BB_1 - OB_1}{BB_1} + \frac{CC_1 - OC_1}{CC_1} = 2$.</p><p>$\left(1 - \frac{OA_1}{AA_1}\right) + \left(1 - \frac{OB_1}{BB_1}\right) + \left(1 - \frac{OC_1}{CC_1}\right) = 2$.</p><p>$3 - \left(\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1}\right) = 2$.</p><p>$\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1} = 3 - 2 = 1$.</p><p>Что и требовалось доказать.</p><p>Ответ: Тождества доказаны.</p><p><strong>в)</strong></p><p>Вопрос заключается в том, может ли точка $O$ делить биссектрису, например $AA_1$, пополам. Это означает, что $AO = OA_1$, или $\frac{AO}{OA_1} = 1$.</p><p>Из пункта <strong>а)</strong> мы знаем, что $\frac{AO}{OA_1} = \frac{b+c}{a}$.</p><p>Таким образом, условие $AO = OA_1$ эквивалентно уравнению $\frac{b+c}{a} = 1$, что означает $b+c = a$.</p><p>Однако, для любого невырожденного треугольника должно выполняться неравенство треугольника, согласно которому сумма длин двух любых сторон должна быть больше длины третьей стороны. В нашем случае, $b+c > a$.</p><p>Равенство $b+c = a$ возможно только для вырожденного треугольника, когда точки $A, B, C$ лежат на одной прямой, и точка $A$ лежит между $B$ и $C$. Но это уже не треугольник.</p><p>Поскольку для любого треугольника $b+c > a$, то $\frac{b+c}{a} > 1$. Следовательно, $\frac{AO}{OA_1} > 1$, то есть $AO > OA_1$.</p><p>Аналогичные рассуждения верны для двух других биссектрис:$\frac{BO}{OB_1} = \frac{a+c}{b} > 1$</p><p>$\frac{CO}{OC_1} = \frac{a+b}{c} > 1$</p><p>Таким образом, точка пересечения биссектрис никогда не может делить биссектрису пополам. Она всегда расположена дальше от вершины, чем от стороны.</p><p>Ответ: Нет, не может.</p><p><strong>г)</strong></p><p>Нам нужно доказать утверждение "тогда и только тогда", что требует доказательства в обе стороны.</p><p>Утверждение: "Одна из биссектрис делится точкой $O$ в отношении 2:1, считая от вершины" $\iff$ "Одна из сторон треугольника равна полусумме двух других сторон".</p><p><strong>Доказательство ($\Rightarrow$):</strong></p><p>Предположим, что одна из биссектрис, например $AA_1$, делится точкой $O$ в отношении 2:1, считая от вершины. Это означает, что $\frac{AO}{OA_1} = \frac{2}{1} = 2$.</p><p>Из пункта <strong>а)</strong> мы знаем, что $\frac{AO}{OA_1} = \frac{b+c}{a}$.</p><p>Приравнивая эти два выражения, получаем:</p><p>$\frac{b+c}{a} = 2$</p><p>$b+c = 2a$</p><p>$a = \frac{b+c}{2}$</p><p>Это означает, что сторона $a$ (противолежащая вершине $A$, из которой выходит биссектриса) равна полусумме двух других сторон $b$ и $c$.Если бы мы предположили, что $\frac{BO}{OB_1}=2$, то получили бы $b = \frac{a+c}{2}$. Если $\frac{CO}{OC_1}=2$, то $c = \frac{a+b}{2}$.Таким образом, первая часть доказана.</p><p><strong>Доказательство ($\Leftarrow$):</strong></p><p>Теперь предположим, что одна из сторон треугольника равна полусумме двух других. Пусть, например, $a = \frac{b+c}{2}$.</p><p>Мы хотим доказать, что одна из биссектрис делится точкой $O$ в отношении 2:1. Найдем отношение, в котором точка $O$ делит биссектрису $AA_1$. Это отношение равно $\frac{AO}{OA_1}$.</p><p>Из пункта <strong>а)</strong> мы знаем формулу: $\frac{AO}{OA_1} = \frac{b+c}{a}$.</p><p>Подставим в эту формулу наше предположение $a = \frac{b+c}{2}$. Из него следует, что $b+c = 2a$.</p><p>$\frac{AO}{OA_1} = \frac{2a}{a} = 2$.</p><p>Это означает, что биссектриса $AA_1$ делится точкой $O$ в отношении 2:1, считая от вершины $A$.Аналогично, если бы мы начали с $b=\frac{a+c}{2}$, то получили бы $\frac{BO}{OB_1}=2$. Если бы $c=\frac{a+b}{2}$, то $\frac{CO}{OC_1}=2$.Таким образом, вторая часть доказана.</p><p>Мы доказали утверждение в обе стороны.</p><p>Ответ: Утверждение доказано.</p>" ] #original: array:6 [ "id" => 1352566 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1139} "task" => array:2 [ "refs" => "745253" "type" => "task" ] "text" => "<p><strong>а)</strong></p><p>Рассмотрим треугольник $ABC$ со сторонами $AB=c$, $BC=a$, $CA=b$. Биссектрисы $AA_1, BB_1, CC_1$ пересекаются в точке $O$.</p><p>Чтобы найти отношение $\frac{AO}{OA_1}$, рассмотрим треугольник $AB B_1$. В этом треугольнике $AO$ является биссектрисой угла $A$. По свойству биссектрисы, она делит противолежащую сторону $BB_1$ в отношении, равном отношению прилежащих сторон:</p><p>$\frac{AO}{OB_1} = \frac{AB}{AB_1}$. Ой, это не то отношение.</p><p>Воспользуемся другим подходом. Рассмотрим треугольник $ACA_1$. В нем $CO$ является биссектрисой угла $C$. По свойству биссектрисы, она делит противолежащую сторону $AA_1$ в отношении:</p><p>$\frac{AO}{OA_1} = \frac{AC}{CA_1}$</p><p>Теперь найдем длину отрезка $CA_1$. $AA_1$ – биссектриса угла $A$ в треугольнике $ABC$. По свойству биссектрисы треугольника:</p><p>$\frac{BA_1}{CA_1} = \frac{AB}{AC} = \frac{c}{b}$</p><p>Мы знаем, что $BA_1 + CA_1 = BC = a$. Из пропорции имеем $BA_1 = \frac{c}{b} CA_1$. Подставим в сумму:</p><p>$\frac{c}{b} CA_1 + CA_1 = a$</p><p>$CA_1 \left(\frac{c}{b} + 1\right) = a$</p><p>$CA_1 \left(\frac{c+b}{b}\right) = a$</p><p>$CA_1 = \frac{ab}{b+c}$</p><p>Теперь подставим $CA_1$ в формулу для искомого отношения:</p><p>$\frac{AO}{OA_1} = \frac{AC}{CA_1} = \frac{b}{\frac{ab}{b+c}} = \frac{b(b+c)}{ab} = \frac{b+c}{a}$</p><p>Аналогично, применяя свойство биссектрисы к другим треугольникам, можно получить остальные отношения.Рассмотрим треугольник $BCB_1$, в нем $AO$ — биссектриса угла $A$. Нет, это неверно.</p><p>Рассмотрим треугольник $ABB_1$. $AO$ является биссектрисой угла $A$. Тогда $\frac{BO}{OB_1} = \frac{AB}{AB_1}$. Найдем $AB_1$. По свойству биссектрисы $BB_1$ в треугольнике $ABC$: $\frac{AB_1}{CB_1} = \frac{AB}{BC} = \frac{c}{a}$. $AB_1 + CB_1 = AC = b$. Отсюда $CB_1 = b - AB_1$.$\frac{AB_1}{b-AB_1} = \frac{c}{a} \Rightarrow a \cdot AB_1 = c(b-AB_1) \Rightarrow (a+c)AB_1 = bc \Rightarrow AB_1 = \frac{bc}{a+c}$.Тогда $\frac{BO}{OB_1} = \frac{c}{\frac{bc}{a+c}} = \frac{a+c}{b}$.</p><p>Рассмотрим треугольник $ACC_1$. $BO$ не является биссектрисой. Рассмотрим треугольник $BCC_1$. $BO$ - биссектриса угла $B$. Тогда $\frac{CO}{OC_1} = \frac{BC}{BC_1}$. Найдем $BC_1$. По свойству биссектрисы $CC_1$ в треугольнике $ABC$: $\frac{AC_1}{BC_1} = \frac{AC}{BC} = \frac{b}{a}$. $AC_1 + BC_1 = AB = c$. Отсюда $AC_1 = c - BC_1$.$\frac{c-BC_1}{BC_1} = \frac{b}{a} \Rightarrow a(c-BC_1) = b \cdot BC_1 \Rightarrow ac = (a+b)BC_1 \Rightarrow BC_1 = \frac{ac}{a+b}$.Тогда $\frac{CO}{OC_1} = \frac{a}{\frac{ac}{a+b}} = \frac{a+b}{c}$.</p><p>Ответ: $\frac{AO}{OA_1} = \frac{b+c}{a}$, $\frac{BO}{OB_1} = \frac{a+c}{b}$, $\frac{CO}{OC_1} = \frac{a+b}{c}$.</p><p><strong>б)</strong></p><p><strong>1. Докажем, что $\frac{AO}{AA_1} + \frac{BO}{BB_1} + \frac{CO}{CC_1} = 2$.</strong></p><p>Используем отношения, найденные в пункте <strong>а)</strong>.Из $\frac{AO}{OA_1} = \frac{b+c}{a}$ следует, что $OA_1 = AO \cdot \frac{a}{b+c}$.Поскольку $AA_1 = AO + OA_1$, то $AA_1 = AO + AO \cdot \frac{a}{b+c} = AO \left(1 + \frac{a}{b+c}\right) = AO \left(\frac{b+c+a}{b+c}\right)$.Отсюда получаем отношение $\frac{AO}{AA_1} = \frac{b+c}{a+b+c}$.</p><p>Аналогично для двух других биссектрис:$\frac{BO}{BB_1} = \frac{a+c}{a+b+c}$</p><p>$\frac{CO}{CC_1} = \frac{a+b}{a+b+c}$</p><p>Сложим эти три отношения:</p><p>$\frac{AO}{AA_1} + \frac{BO}{BB_1} + \frac{CO}{CC_1} = \frac{b+c}{a+b+c} + \frac{a+c}{a+b+c} + \frac{a+b}{a+b+c} = \frac{(b+c) + (a+c) + (a+b)}{a+b+c} = \frac{2a+2b+2c}{a+b+c} = \frac{2(a+b+c)}{a+b+c} = 2$.</p><p>Что и требовалось доказать.</p><p><strong>2. Докажем, что $\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1} = 1$.</strong>(Предполагается, что в условии опечатка и имеется в виду $OA_1$, а не $AO_1$).</p><p>Это можно доказать двумя способами.</p><p>Способ 1. Выразим отношение $\frac{OA_1}{AA_1}$.Так как $AA_1 = AO + OA_1$, то $\frac{OA_1}{AA_1} = \frac{OA_1}{AO+OA_1}$. Разделим числитель и знаменатель на $OA_1$:</p><p>$\frac{OA_1}{AA_1} = \frac{1}{\frac{AO}{OA_1} + 1}$.</p><p>Используя результат из пункта <strong>а)</strong> $\frac{AO}{OA_1} = \frac{b+c}{a}$:</p><p>$\frac{OA_1}{AA_1} = \frac{1}{\frac{b+c}{a} + 1} = \frac{1}{\frac{b+c+a}{a}} = \frac{a}{a+b+c}$.</p><p>Аналогично: $\frac{OB_1}{BB_1} = \frac{b}{a+b+c}$ и $\frac{OC_1}{CC_1} = \frac{c}{a+b+c}$.</p><p>Сложим эти три отношения:</p><p>$\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1} = \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c} = \frac{a+b+c}{a+b+c} = 1$.</p><p>Что и требовалось доказать.</p><p>Способ 2. Используем уже доказанное тождество.</p><p>$\frac{AO}{AA_1} + \frac{BO}{BB_1} + \frac{CO}{CC_1} = 2$.</p><p>Заменим $AO = AA_1 - OA_1$, $BO = BB_1 - OB_1$, $CO = CC_1 - OC_1$:</p><p>$\frac{AA_1 - OA_1}{AA_1} + \frac{BB_1 - OB_1}{BB_1} + \frac{CC_1 - OC_1}{CC_1} = 2$.</p><p>$\left(1 - \frac{OA_1}{AA_1}\right) + \left(1 - \frac{OB_1}{BB_1}\right) + \left(1 - \frac{OC_1}{CC_1}\right) = 2$.</p><p>$3 - \left(\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1}\right) = 2$.</p><p>$\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1} = 3 - 2 = 1$.</p><p>Что и требовалось доказать.</p><p>Ответ: Тождества доказаны.</p><p><strong>в)</strong></p><p>Вопрос заключается в том, может ли точка $O$ делить биссектрису, например $AA_1$, пополам. Это означает, что $AO = OA_1$, или $\frac{AO}{OA_1} = 1$.</p><p>Из пункта <strong>а)</strong> мы знаем, что $\frac{AO}{OA_1} = \frac{b+c}{a}$.</p><p>Таким образом, условие $AO = OA_1$ эквивалентно уравнению $\frac{b+c}{a} = 1$, что означает $b+c = a$.</p><p>Однако, для любого невырожденного треугольника должно выполняться неравенство треугольника, согласно которому сумма длин двух любых сторон должна быть больше длины третьей стороны. В нашем случае, $b+c > a$.</p><p>Равенство $b+c = a$ возможно только для вырожденного треугольника, когда точки $A, B, C$ лежат на одной прямой, и точка $A$ лежит между $B$ и $C$. Но это уже не треугольник.</p><p>Поскольку для любого треугольника $b+c > a$, то $\frac{b+c}{a} > 1$. Следовательно, $\frac{AO}{OA_1} > 1$, то есть $AO > OA_1$.</p><p>Аналогичные рассуждения верны для двух других биссектрис:$\frac{BO}{OB_1} = \frac{a+c}{b} > 1$</p><p>$\frac{CO}{OC_1} = \frac{a+b}{c} > 1$</p><p>Таким образом, точка пересечения биссектрис никогда не может делить биссектрису пополам. Она всегда расположена дальше от вершины, чем от стороны.</p><p>Ответ: Нет, не может.</p><p><strong>г)</strong></p><p>Нам нужно доказать утверждение "тогда и только тогда", что требует доказательства в обе стороны.</p><p>Утверждение: "Одна из биссектрис делится точкой $O$ в отношении 2:1, считая от вершины" $\iff$ "Одна из сторон треугольника равна полусумме двух других сторон".</p><p><strong>Доказательство ($\Rightarrow$):</strong></p><p>Предположим, что одна из биссектрис, например $AA_1$, делится точкой $O$ в отношении 2:1, считая от вершины. Это означает, что $\frac{AO}{OA_1} = \frac{2}{1} = 2$.</p><p>Из пункта <strong>а)</strong> мы знаем, что $\frac{AO}{OA_1} = \frac{b+c}{a}$.</p><p>Приравнивая эти два выражения, получаем:</p><p>$\frac{b+c}{a} = 2$</p><p>$b+c = 2a$</p><p>$a = \frac{b+c}{2}$</p><p>Это означает, что сторона $a$ (противолежащая вершине $A$, из которой выходит биссектриса) равна полусумме двух других сторон $b$ и $c$.Если бы мы предположили, что $\frac{BO}{OB_1}=2$, то получили бы $b = \frac{a+c}{2}$. Если $\frac{CO}{OC_1}=2$, то $c = \frac{a+b}{2}$.Таким образом, первая часть доказана.</p><p><strong>Доказательство ($\Leftarrow$):</strong></p><p>Теперь предположим, что одна из сторон треугольника равна полусумме двух других. Пусть, например, $a = \frac{b+c}{2}$.</p><p>Мы хотим доказать, что одна из биссектрис делится точкой $O$ в отношении 2:1. Найдем отношение, в котором точка $O$ делит биссектрису $AA_1$. Это отношение равно $\frac{AO}{OA_1}$.</p><p>Из пункта <strong>а)</strong> мы знаем формулу: $\frac{AO}{OA_1} = \frac{b+c}{a}$.</p><p>Подставим в эту формулу наше предположение $a = \frac{b+c}{2}$. Из него следует, что $b+c = 2a$.</p><p>$\frac{AO}{OA_1} = \frac{2a}{a} = 2$.</p><p>Это означает, что биссектриса $AA_1$ делится точкой $O$ в отношении 2:1, считая от вершины $A$.Аналогично, если бы мы начали с $b=\frac{a+c}{2}$, то получили бы $\frac{BO}{OB_1}=2$. Если бы $c=\frac{a+b}{2}$, то $\frac{CO}{OC_1}=2$.Таким образом, вторая часть доказана.</p><p>Мы доказали утверждение в обе стороны.</p><p>Ответ: Утверждение доказано.</p>" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "next" => array:2 [ "refs" => "745254" "type" => "task" ] "previous" => array:2 [ "refs" => "745252" "type" => "task" ] "book" => Illuminate\Database\Eloquent\Collection {#1113 #items: array:1 [ 0 => App\Models\Book {#1050} ] #escapeWhenCastingToString: false } "page" => Illuminate\Database\Eloquent\Collection {#1038 #items: array:1 [ 0 => App\Models\BookPage {#1117 #connection: "mysql" #table: "book_pages" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] 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№838 (с. 212)
Условие. №838 (с. 212)
скриншот условия
838. Биссектрисы АА₁, ВВ₁ и СС₁ треугольника АВС со сторонами АВ = с, ВС = а и СА = b пересекаются в точке О.
в) Может ли хотя бы одна из биссектрис треугольника делиться точкой О пополам? г) Докажите, что одна из биссектрис делится точкой О в отношении 2 : 1, считая от вершины, тогда и только тогда, когда одна из сторон треугольника равна полусумме двух других сторон.
Решение 2. №838 (с. 212)
Решение 3. №838 (с. 212)
Решение 6. №838 (с. 212)
а)
Рассмотрим треугольник $ABC$ со сторонами $AB=c$, $BC=a$, $CA=b$. Биссектрисы $AA_1, BB_1, CC_1$ пересекаются в точке $O$.
Чтобы найти отношение $\frac{AO}{OA_1}$, рассмотрим треугольник $AB B_1$. В этом треугольнике $AO$ является биссектрисой угла $A$. По свойству биссектрисы, она делит противолежащую сторону $BB_1$ в отношении, равном отношению прилежащих сторон:
$\frac{AO}{OB_1} = \frac{AB}{AB_1}$. Ой, это не то отношение.
Воспользуемся другим подходом. Рассмотрим треугольник $ACA_1$. В нем $CO$ является биссектрисой угла $C$. По свойству биссектрисы, она делит противолежащую сторону $AA_1$ в отношении:
$\frac{AO}{OA_1} = \frac{AC}{CA_1}$
Теперь найдем длину отрезка $CA_1$. $AA_1$ – биссектриса угла $A$ в треугольнике $ABC$. По свойству биссектрисы треугольника:
$\frac{BA_1}{CA_1} = \frac{AB}{AC} = \frac{c}{b}$
Мы знаем, что $BA_1 + CA_1 = BC = a$. Из пропорции имеем $BA_1 = \frac{c}{b} CA_1$. Подставим в сумму:
$\frac{c}{b} CA_1 + CA_1 = a$
$CA_1 \left(\frac{c}{b} + 1\right) = a$
$CA_1 \left(\frac{c+b}{b}\right) = a$
$CA_1 = \frac{ab}{b+c}$
Теперь подставим $CA_1$ в формулу для искомого отношения:
$\frac{AO}{OA_1} = \frac{AC}{CA_1} = \frac{b}{\frac{ab}{b+c}} = \frac{b(b+c)}{ab} = \frac{b+c}{a}$
Аналогично, применяя свойство биссектрисы к другим треугольникам, можно получить остальные отношения.Рассмотрим треугольник $BCB_1$, в нем $AO$ — биссектриса угла $A$. Нет, это неверно.
Рассмотрим треугольник $ABB_1$. $AO$ является биссектрисой угла $A$. Тогда $\frac{BO}{OB_1} = \frac{AB}{AB_1}$. Найдем $AB_1$. По свойству биссектрисы $BB_1$ в треугольнике $ABC$: $\frac{AB_1}{CB_1} = \frac{AB}{BC} = \frac{c}{a}$. $AB_1 + CB_1 = AC = b$. Отсюда $CB_1 = b - AB_1$.$\frac{AB_1}{b-AB_1} = \frac{c}{a} \Rightarrow a \cdot AB_1 = c(b-AB_1) \Rightarrow (a+c)AB_1 = bc \Rightarrow AB_1 = \frac{bc}{a+c}$.Тогда $\frac{BO}{OB_1} = \frac{c}{\frac{bc}{a+c}} = \frac{a+c}{b}$.
Рассмотрим треугольник $ACC_1$. $BO$ не является биссектрисой. Рассмотрим треугольник $BCC_1$. $BO$ - биссектриса угла $B$. Тогда $\frac{CO}{OC_1} = \frac{BC}{BC_1}$. Найдем $BC_1$. По свойству биссектрисы $CC_1$ в треугольнике $ABC$: $\frac{AC_1}{BC_1} = \frac{AC}{BC} = \frac{b}{a}$. $AC_1 + BC_1 = AB = c$. Отсюда $AC_1 = c - BC_1$.$\frac{c-BC_1}{BC_1} = \frac{b}{a} \Rightarrow a(c-BC_1) = b \cdot BC_1 \Rightarrow ac = (a+b)BC_1 \Rightarrow BC_1 = \frac{ac}{a+b}$.Тогда $\frac{CO}{OC_1} = \frac{a}{\frac{ac}{a+b}} = \frac{a+b}{c}$.
Ответ: $\frac{AO}{OA_1} = \frac{b+c}{a}$, $\frac{BO}{OB_1} = \frac{a+c}{b}$, $\frac{CO}{OC_1} = \frac{a+b}{c}$.
б)
1. Докажем, что $\frac{AO}{AA_1} + \frac{BO}{BB_1} + \frac{CO}{CC_1} = 2$.
Используем отношения, найденные в пункте а).Из $\frac{AO}{OA_1} = \frac{b+c}{a}$ следует, что $OA_1 = AO \cdot \frac{a}{b+c}$.Поскольку $AA_1 = AO + OA_1$, то $AA_1 = AO + AO \cdot \frac{a}{b+c} = AO \left(1 + \frac{a}{b+c}\right) = AO \left(\frac{b+c+a}{b+c}\right)$.Отсюда получаем отношение $\frac{AO}{AA_1} = \frac{b+c}{a+b+c}$.
Аналогично для двух других биссектрис:$\frac{BO}{BB_1} = \frac{a+c}{a+b+c}$
$\frac{CO}{CC_1} = \frac{a+b}{a+b+c}$
Сложим эти три отношения:
$\frac{AO}{AA_1} + \frac{BO}{BB_1} + \frac{CO}{CC_1} = \frac{b+c}{a+b+c} + \frac{a+c}{a+b+c} + \frac{a+b}{a+b+c} = \frac{(b+c) + (a+c) + (a+b)}{a+b+c} = \frac{2a+2b+2c}{a+b+c} = \frac{2(a+b+c)}{a+b+c} = 2$.
Что и требовалось доказать.
2. Докажем, что $\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1} = 1$.(Предполагается, что в условии опечатка и имеется в виду $OA_1$, а не $AO_1$).
Это можно доказать двумя способами.
Способ 1. Выразим отношение $\frac{OA_1}{AA_1}$.Так как $AA_1 = AO + OA_1$, то $\frac{OA_1}{AA_1} = \frac{OA_1}{AO+OA_1}$. Разделим числитель и знаменатель на $OA_1$:
$\frac{OA_1}{AA_1} = \frac{1}{\frac{AO}{OA_1} + 1}$.
Используя результат из пункта а) $\frac{AO}{OA_1} = \frac{b+c}{a}$:
$\frac{OA_1}{AA_1} = \frac{1}{\frac{b+c}{a} + 1} = \frac{1}{\frac{b+c+a}{a}} = \frac{a}{a+b+c}$.
Аналогично: $\frac{OB_1}{BB_1} = \frac{b}{a+b+c}$ и $\frac{OC_1}{CC_1} = \frac{c}{a+b+c}$.
Сложим эти три отношения:
$\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1} = \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c} = \frac{a+b+c}{a+b+c} = 1$.
Что и требовалось доказать.
Способ 2. Используем уже доказанное тождество.
$\frac{AO}{AA_1} + \frac{BO}{BB_1} + \frac{CO}{CC_1} = 2$.
Заменим $AO = AA_1 - OA_1$, $BO = BB_1 - OB_1$, $CO = CC_1 - OC_1$:
$\frac{AA_1 - OA_1}{AA_1} + \frac{BB_1 - OB_1}{BB_1} + \frac{CC_1 - OC_1}{CC_1} = 2$.
$\left(1 - \frac{OA_1}{AA_1}\right) + \left(1 - \frac{OB_1}{BB_1}\right) + \left(1 - \frac{OC_1}{CC_1}\right) = 2$.
$3 - \left(\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1}\right) = 2$.
$\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1} = 3 - 2 = 1$.
Что и требовалось доказать.
Ответ: Тождества доказаны.
в)
Вопрос заключается в том, может ли точка $O$ делить биссектрису, например $AA_1$, пополам. Это означает, что $AO = OA_1$, или $\frac{AO}{OA_1} = 1$.
Из пункта а) мы знаем, что $\frac{AO}{OA_1} = \frac{b+c}{a}$.
Таким образом, условие $AO = OA_1$ эквивалентно уравнению $\frac{b+c}{a} = 1$, что означает $b+c = a$.
Однако, для любого невырожденного треугольника должно выполняться неравенство треугольника, согласно которому сумма длин двух любых сторон должна быть больше длины третьей стороны. В нашем случае, $b+c > a$.
Равенство $b+c = a$ возможно только для вырожденного треугольника, когда точки $A, B, C$ лежат на одной прямой, и точка $A$ лежит между $B$ и $C$. Но это уже не треугольник.
Поскольку для любого треугольника $b+c > a$, то $\frac{b+c}{a} > 1$. Следовательно, $\frac{AO}{OA_1} > 1$, то есть $AO > OA_1$.
Аналогичные рассуждения верны для двух других биссектрис:$\frac{BO}{OB_1} = \frac{a+c}{b} > 1$
$\frac{CO}{OC_1} = \frac{a+b}{c} > 1$
Таким образом, точка пересечения биссектрис никогда не может делить биссектрису пополам. Она всегда расположена дальше от вершины, чем от стороны.
Ответ: Нет, не может.
г)
Нам нужно доказать утверждение "тогда и только тогда", что требует доказательства в обе стороны.
Утверждение: "Одна из биссектрис делится точкой $O$ в отношении 2:1, считая от вершины" $\iff$ "Одна из сторон треугольника равна полусумме двух других сторон".
Доказательство ($\Rightarrow$):
Предположим, что одна из биссектрис, например $AA_1$, делится точкой $O$ в отношении 2:1, считая от вершины. Это означает, что $\frac{AO}{OA_1} = \frac{2}{1} = 2$.
Из пункта а) мы знаем, что $\frac{AO}{OA_1} = \frac{b+c}{a}$.
Приравнивая эти два выражения, получаем:
$\frac{b+c}{a} = 2$
$b+c = 2a$
$a = \frac{b+c}{2}$
Это означает, что сторона $a$ (противолежащая вершине $A$, из которой выходит биссектриса) равна полусумме двух других сторон $b$ и $c$.Если бы мы предположили, что $\frac{BO}{OB_1}=2$, то получили бы $b = \frac{a+c}{2}$. Если $\frac{CO}{OC_1}=2$, то $c = \frac{a+b}{2}$.Таким образом, первая часть доказана.
Доказательство ($\Leftarrow$):
Теперь предположим, что одна из сторон треугольника равна полусумме двух других. Пусть, например, $a = \frac{b+c}{2}$.
Мы хотим доказать, что одна из биссектрис делится точкой $O$ в отношении 2:1. Найдем отношение, в котором точка $O$ делит биссектрису $AA_1$. Это отношение равно $\frac{AO}{OA_1}$.
Из пункта а) мы знаем формулу: $\frac{AO}{OA_1} = \frac{b+c}{a}$.
Подставим в эту формулу наше предположение $a = \frac{b+c}{2}$. Из него следует, что $b+c = 2a$.
$\frac{AO}{OA_1} = \frac{2a}{a} = 2$.
Это означает, что биссектриса $AA_1$ делится точкой $O$ в отношении 2:1, считая от вершины $A$.Аналогично, если бы мы начали с $b=\frac{a+c}{2}$, то получили бы $\frac{BO}{OB_1}=2$. Если бы $c=\frac{a+b}{2}$, то $\frac{CO}{OC_1}=2$.Таким образом, вторая часть доказана.
Мы доказали утверждение в обе стороны.
Ответ: Утверждение доказано.
Другие задания:
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