We are given an acute-angled triangle $ABC$ with $AB < AC < BC$. Points $K$ and $L$ are chosen on segments $AC$ and $BC$, respectively, so that $AB = CK = CL$. Perpendicular bisectors of segments $AK$ and $BL$ intersect the line $AB$ at points $P$ and $Q$, respectively. Segments $KP$ and $LQ$ intersect at point $M$. Prove that $AK + KM = BL + LM$. Poland