Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Wednesday, February 18, 2015
Geometry Problem 1085: Intersecting Circles, Concyclic Points, Cyclic Quadrilateral
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∠GAH
ReplyDelete= ∠CAD
= ∠ADF − ∠ACE
= ∠ABF − ∠ABE
= ∠EBF
= ∠GBH
Hence, ABHG are concyclic.
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Corollary
Since ∠GHB = ∠DAB = ∠DFB
Thus, GH // CF.
Denote ∠DAB = ∠DFB = x and ∠ABE = ∠ACE = y
ReplyDeleteClearly ∠G = x - y = ∠H and so A,G,H,B are concyclic
< ABF = <ADF......(1)
ReplyDelete<ABE = <ACF .....,(2)
(1)-(2) < EBF = < CAD = < GAH
so ABHG is concylic
Sumith Peiris
Moratuwa
Sri Lanka
It also follows that CF and GH are parallel
ReplyDelete