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Thread: triangle

  1. #1
    Senior Member
    Jul 2008

    Exclamation triangle

    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?
    Last edited by xwrathbringerx; Aug 29th 2008 at 03:38 AM.
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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    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.
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