Продлеваем C D и D G до пересечения в точке Z.
Рассматриваем три треугольника с вершиной Z.
Для двух не прямоугольных треугольников вспоминаем теорему косинусов (для угла Z), а для прямоугольного - косинус угла Z через гипотенузу и катет.
Получаем систему трех уравнений, из которой с легкостью можно получить систему двух уравнений третьей степени. Дальше я не берусь решать.
А потом подумал, еще раз подумал и решил. 😁 😜
Формулы написать?
Да! Хотелось бы решение посмотреть.
Продлеваем C D и D G до пересечения в точке Z.
Вспомним теорему косинусов.
a^2 = b^2 + c^2 - 2 * cos(A)
Тогда:
A^2 = K^2 + L^2 -2 * cos(Z)
И
E^2 = (K + B + C)^2 + (L + D + G)^ - 2 * cos(Z)
для прямоугольного - косинус угла Z через гипотенузу и катет:
cos(Z) = (L + D) / (K + 😎
Вот, пожалуй, и все.
Осталось решить систему из двух уравнений.
Найдем К и L, все остальное находится элементарно.
PS
Глюки смайлграфии 😎 следует читать как “В)”
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