triangle

xwrathbringerx · #1 $Old$ 08-29-2008, 04:20 AM

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?

red_dog · #2 $Old$ 08-29-2008, 04:12 PM

We have $\widehat{KBP}=\widehat{KML}$ and $\widehat{KPB}=\widehat{KML}$
Then $\widehat{KBP}=\widehat{KPB}\Rightarrow KP=KB=AK$ .
So, in triangle APB the median PK is half of the side AB. Then the triangle is right-angled in P.