Question

В треугольнике ABC биссектриса угла А делит высоту,проведенную из вершины В, в отношении 13:12,считая от точки В. Найдите длину стороны ВС треугольника, если радиус описанной около него окружности равен 26 см.задание из 2 части. Хз кто решит:)

Answer 1

BH-высота ,опущенная на АС 
рассмотрим треугольник АВН-прямоугольный 
АВ/ВО=АН/ОН 
О-точка пересечения биссектрисы и высоты 
АН=60х 
АВ=65х 
sin(BAH)=BH/AB=7/13 
BC/sin(BAH)=2R-теорема синусов 
ВС=52*7/13 
ВС=28 см 

Answer 2

I found some web search results that might help you with your question. Here is a summary of the answer:

- In the triangle ABC, the bisector of angle A divides the height drawn from vertex B in the ratio 13:12, counting from point B. Find the length of side BC of the triangle, if the radius of the circle circumscribed around it is 26 cm. - The bisector in a triangle divides the side that it intersects in the ratio of the adjacent sides. Therefore, AB:AN = 13:12. From this ratio of the hypotenuse to the cathetus, it is clear that the sides of the right triangle ABN are related to the so-called Pythagorean triples (13, 12, 5). We check by the Pythagorean theorem. Let AB = 13x, AN = 12x, then BN = √(AB² - AN²) = √(169x² - 144x²) = 5x. Then sin A = BH:AB = 5:13. By the sine theorem, BC:sin A = 2R. BC:(5/13) = 2R. 13BC = 260 cm. BC = 20 cm.

You can find more details and a diagram in the first web search result. I hope this helps you.