Challenging Geometry Puzzle: Problem 1555. 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 a = < ATC = < ABT (tangent & alternate segment)

ReplyDeleteLet b = < BTD = < BAT (tangent & alternate segment)

Let < TBD = u

< TDE = u + b = < CAB (since ABDC is concyclic)

But < CAB = < CAT + < BAT

So u + b = < CAT + b

Hence < CAT = u

Now < TBD = < TDE - < BTD = (u + b) - b = u

Hence in quadrilateral ABEF, HAE = < HBF = u so it is concylic

So < FEH = < FAB = < CDE ( = u + b)

Hence EF // CD

Therefore DE / FC = EH / FH and so

DE / 2 = 4 / 6

which gives DE = 4/3

Sumith Peiris

Moratuwa

Sri Lanka