http://www.mathhelpforum.com/math-he...cture43-io.bmp

In the acute-angled triangle ABC, K is the midpoint of AB, L is the midpoint of BC and M is the midpoint of CA. The circle through K, L and M also cuts BC at P as shown in the diagram. KMLB is a parallelogram and angle KPB is equal to KML. Prove that AP is perpendicular to BC. Can anyone get me started?