IMO 2012 Problem 5

Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$ , and let $D$ be the foot of the altitude from $C$ . Let $X$ be a point in the interior of the segment $CD$ . Let $K$ be the point on the segment $AX$ such that $BK=BC$ . Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$ . Let $M$ be the point of intersection of $AL$ and $BK$ .

Show that $MK=ML$ .

Proposed by Josef Tkadlec, Czech Republic