Problem

Source: RMO KV 2024 Q5

Tags: geometry



Let $ABC$ be a triangle with $\angle ABC = 20^{\circ}$ and $\angle ACB = 40^{\circ}$. Let $D$ be a point on $BC$ such that $\angle BAD = \angle DAC$. Let the incircle of triangle $ABC$ touch $BC$ at $E$. Prove that $BD = 2 \cdot CE$.