Предмет: Геометрия, автор: Diana778

Помогите пожалуйста) Задания в фото.

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Автор ответа: alimhanchik
0
1 и 2 решается через равенство треугольников:

1)MDK и PDE - равные треуг-ки по 2-ум сторонам и углу межу ними (MD=DE, PD=DK -тк делятся пополам и углы PDE и MDK - вертикальные)
Отсюда и искомые углы равны, ибо треугольники равны, понятно и так.

2) Тр-ки KDP и DPM равны по трем сторонам: кд=дм,кр=рм - по условию, а др-общая. Сл-но углы кдр и рдм равны, а это значит, что др - биссектриса. Чтд.

3) Пусть х-1 часть, то основание = 2х, а тк это равноб трк, то боковые стороны равны попарно по 3х, тогда составляю ур-е:
3х+3х+2х=56
8х=56
Х=7
Сл-но основание = 14
Боковые стороны по 21.
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