Let ABC be an acute triangle with AC>AB>BC. The perpendicular bisectors of AC and AB cut line BC at D and E respectively. Let P and Q be points on lines AC and AB respectively, both different from A, such that AB=BP and AC=CQ, and let K be the intersection of lines EP and DQ. Let M be the midpoint of BC. Show that ∠DKA=∠EKM. geometryperpendicular bisector