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Old 08-29-2008   #1
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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?
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Old 08-29-2008   #2
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Are you planning to post all your questions in all the existing maths websites ?
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Old 09-19-2008   #3
Math Tyrant
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Default Hmmm......

I wish we could see the diagram for this. Because then it would be easier to answer your question........Maybe if you were to make this diagram and then copied and pasted it this question of yours could be answerd alot faster.
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Old 09-19-2008   #4
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start with your main points
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