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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Let CA and BD extended meet at F. Draw FG // DE, G on BE extended.

ReplyDeleteLet 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

Moratuwa

Sri Lanka

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.

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

ReplyDelete