Challenging Geometry Puzzle: Problem 1582. 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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In Circle Q,
ReplyDeleteLet < ACF = < AEF = x
Let < AFB = y
Let < FAE = < FCE = z
So from Circle O, < ADB = < ACE = x + z
But in Triangle ADF, < ADB = x + y
Hence y = z and
< AFD = < AEF
Sumith Peiris
Moratuwa
Sri Lanka
It follows that AF is tangential to Circle BCF at F
DeleteSumith Peiris
Moratuwa
Sri Lanka
... and to circle (FED) as well!
Delete