Let $t$ be a line passing through the vertex $A$ of the equilateral $ABC$, parallel to the side $BC$. On the side $AC$ arbitrarily mark the point $D$. Bisector of the angle $ABD$ intersects the line $t$at the point $E$. Prove that $BD=CD+AE$.
Problem
Source: Champions Tournament (Ukraine) - Турнір чемпіонів - 2016 Seniors p3
Tags: geometry, equal segments, Equilateral, Champions Tournament