## Wednesday, September 27, 2023

### Geometry Problem 1555: Find Length of DE

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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#### 1 comment:

1. Let a = < ATC = < ABT (tangent & alternate segment)
Let 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