MathNet 1974 — Part 01 Question 9

In the triangle $ABC$, $\angle C = 90^\circ$ and $AC = BC$. Take points $D$ on $CA$ and $E$ on $CB$ such that $CD = CE$. Let the perpendiculars from $D$ and $C$ to $AE$ meet $AB$ at $K$ and $L$ respectively. Show that $KL = LB$.