Saturday, July 13, 2024

Geometry Problem 1562: Collinearity in a Right-angled Triangle involving the Altitude, Angle bisectors, and Midpoint.

Challenging Geometry Puzzle: Problem 1562. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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Illustration of Geometry Problem 1562

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Posted by Antonio Gutierrez at 1:00 PM
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2 comments:

  1. Georgios Kousinioris, Gastouni GreeceJuly 18, 2024 at 11:51 PM

    Let EF intersects BD at N
    <EHF=<EBF=90° thus B, E, H, F concyclic
    <BEF=<BHF=45°=<EBD hence NE=NB.
    Similarly NB=NF so NE=NF therefore N is the midpoint EF
    <BDC=<A+<B/2=45°+<A = <BEH thus B, E, D, F concyclic
    <EDF=<EHF=90° and DN median so DN=EF/2=BN
    Therefore N is the midpoint BD

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  2. Sumith PeirisJuly 23, 2024 at 10:20 AM

    Let EF, BD cut at K

    < EBD = 45 = < EHD, hence B,E,D,H,F is concyclic and < BDE = < BHE = 45 = < DBE

    It follows that the 6 triangles BDK, BFK, BEF, DKE, DKF, DEF are all right isosceles and so BEDF is a square and BK = KD

    But BM = MD so M and K coincide

    Therefore E,M,F are collinear

    Sumith Peiris
    Moratuwa
    Sri Lanka

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