Problem

Source: Problem 4

Tags: geometry, Angle Chasing, similar triangles



Let $ABC$ be a triangle with $\angle A=90^{\circ}$ and $AB=AC$. Let $D$ and $E$ be points on the segment $BC$ such that $BD:DE:EC = 1:2:\sqrt{3}$. Prove that $\angle DAE= 45^{\circ}$