# triangle

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• August 29th 2008, 02:20 AM
xwrathbringerx
triangle
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?
• August 29th 2008, 02:12 PM
red_dog
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.