Предмет: Геометрия, автор: Legend7

докажите что биссектриса любого из внешних углов при вершине равнобедренного треугольника параллельна основанию.

Ответы

Автор ответа: hantrider
0

KBC – внешний угол,

KBC=  А +  С

KBC= 2С

т.к. ВD – биссектриса
KBD= DBC = С

DBC=С, значит,
BD||АС

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