Thursday, November 2, 2023

Geometry Problem 1559: Proving BC Bisects Segment DE

Challenging Geometry Puzzle: Problem 1559. 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 1559: Proving BC Bisects Segment DE

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  1. Let CA and BD extended meet at F. Draw FG // DE, G on BE extended.
    Let FG and BC cut at N. Extend AC to L

    Let FD = u, DB = v, AD = w
    Let BE = x, EG = y , CE = z

    Since DE//FG, v/u = x/y........(1)

    Since Triangles ABD and BCE are similar
    v/w = x/z...............(2)

    From (1) and (2) x/v = y/u = z/w ..........(3)

    Now in Triangles ADF & CEG from (3)
    u/w = y/z and the included angles < ADF = < CEG (= alpha+beta)
    Hence Triangles ADF & CEG are similar & < AFD = < CGE = < GCL since CF // BG

    So BFCG is a parallelogram and the diagonals bisect each other

    Hence BN bisects FG and since FG//DE, BN (or BM) also bisects DE

    Sumith Peiris
    Sri Lanka

    1. Draw line DN //BE and connect E TO N on INTERSECTION DN TO BC . And INTERSECTION BA AND DN is S. We can prove BEND is parallelogram BE=X BD=V EN=L EC=S DA=K NC=R SA=P Triangles BDA AND BEC are similar so X/V=S/K=R/P In Triangle BAC R/P=T/U SO T/U=R/P=S/K=X/V it prove that Triangles BEN and BDS are similar & <BDN=<BEN SOBEDN IS parallelogram and the diagonals bisect each other. BEHROUZ L.A.

  2. I think the problem would have been more difficult if it required to prove that AB is symmedian of triangle BDE