In the triangle ABC, the bisector BE and the median AD are perpendicular and have

In the triangle ABC, the bisector BE and the median AD are perpendicular and have the same length equal to 168. Find the sides of the triangle ABC.

Let’s denote the intersection of AD and BE by point O. Then in the triangle ABD, BО will be the height, and ADB – isosceles, AB = BD = DC, 2AB = BC. Since the bisector divides AC as both the AB / BC ratio, AC = 3AE. Let’s draw a line through the vertex B parallel to AC until it intersects with the extension AD, the intersection point is denoted as M. Then BM = AC (because ABMC is a parallelogram). Both AOE and BOM are similar (AOE = BOM = 90, EAO = AMD, like angles on parallel lines). And BM / AE = 3, and BO / OE = 3, OE = 42. AE = √ (AO ^ 2 + OE ^ 2) = 42√5, AC = 3 * 42√5 = 126√5, AB = 42√13, BC = 2AB = 84√13