Предмет: Геометрия, автор: ed2009

В треугольнике АВС АС=СВ , СК - биссектриса , АВ = 10 см , СК = 5 см. Найдите угол А
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Автор ответа: korchmit

2

В равнобедренном треугольнике биссектриса угла, лежащего против основания, является медианой и высотой.
Значит СК делит сторону АВ пополам и перпендикулярна ей. Отсюда АК равна 5 см, т.е. треугольник АСК прямоугольный равнобедренный. Следовательно угол А равен 45°

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