D is a point on the edge BC of triangle ABC such that AD\equal{}\frac{BD^2}{AB\plus{}AD}\equal{}\frac{CD^2}{AC\plus{}AD}. E is a point such that D is on [AE] and CD\equal{}\frac{DE^2}{CD\plus{}CE}. Prove that AE\equal{}AB\plus{}AC. trigonometrygeometryangle bisectorgeometry unsolved