2012 IMO Problems/Problem 4

First we construct three circles:the circumcircles of ABC() ,The circle() with centre A and radius AC and the circle( ) with centre B and radius BC. Note that the centre of lies on the midpoint of AB,so the three circles are co-axial with radical axis CD. Let AX=Y=ABX=Z=BAYBZ=P . so AYB=AZB=90,so X must be the orthocenter of ABP implying P lies on the radical axis. let us denote PWR(P) by the power of P wrt .[which are equal] From similar triangles ABCACD we get AC2=ADAB=AL2ALD=ABL similarly BKD=BAK now APD=ABZ=ALZ implying ADLP is cyclic. so ALPL.....(1) meaning PL2=PWR(P). similarly we get BKPK.....(2) meaning PK2=PWR(P) hence PK=PL.....(3) so MKP=MLP[using (1),(2),(3)] proving MK=ML