Results 1 to 2 of 2

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 02:38 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    We have $\displaystyle \widehat{KBP}=\widehat{KML}$ and $\displaystyle \widehat{KPB}=\widehat{KML}$
    Then $\displaystyle \widehat{KBP}=\widehat{KPB}\Rightarrow KP=KB=AK$.
    So, in triangle APB the median PK is half of the side AB. Then the triangle is right-angled in P.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Apr 23rd 2011, 08:10 AM
  2. Replies: 3
    Last Post: Apr 30th 2009, 07:41 AM
  3. Replies: 1
    Last Post: Oct 28th 2008, 07:02 PM
  4. Replies: 7
    Last Post: Jul 19th 2008, 06:53 AM
  5. Replies: 27
    Last Post: Apr 27th 2008, 10:36 AM

Search Tags

/mathhelpforum @mathhelpforum