Friday, June 19, 2015

Geometry Problem 1124: Triangle, Median, Two Squares, Area

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1124: Triangle, Median, Two Squares, Area.

Posted by Antonio Gutierrez at 10:05 AM
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3 comments:

  1. Jacob HA (EMK2000)June 19, 2015 at 2:15 PM

    Consider rotation anti-clockwise 90° about M,
    then D→C and F→B.
    Thus ΔMDF→ΔMCB and hence congruent.

    The result S₁+S₃=S₂+S₄ then follows.

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  2. Sumith PeirisJuly 18, 2015 at 10:58 PM

    Tr.s MBC & MFD are congruent SAS hence they are equal in area and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  3. APOSTOLIS MANOLOUDISSeptember 13, 2016 at 11:05 AM

    Problem 1224
    Let area triangle ABC is (ABC).Then (BMC)=(ABM) (BM=MC). But <BMA+<DMF=180 so
    (MDF)/(ABM)=(MF.MD)/(BM.AM)=1.So (BMC)=(MDF) or S_1+(MHPQ)+S_(3 ) =S_2 +(MHPQ)+S_4.
    Therefore S_1+S_(3 )=S_2+S_4.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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