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

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

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