Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

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## Thursday, July 28, 2016

### Geometry Problem 1239: Intersecting Circles, Secant, Concyclic Points, Congruence

Labels:
concyclic,
congruence,
intersecting circles,
secant

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∠BFG=∠BFC=∠BAD=∠BED

ReplyDelete∴B E F G concyclic

∠EBH=180°-∠EBF=180°-∠EAC=∠EAD

∴ED=EH

nice solution. Thank you very much.

ReplyDeleteProblem 1239

ReplyDeleteIs <CEG=<ECD+<EDC=<BCA+<EBA=<BFA+<CBA=<BFA+<CFA=<CFB=<GFB, so F,G,E and B are concyclic. But <EBH=180-<CBF=180-<CAF=<EAD,then EH=ED.

4 HIGH SCHOOL OF KORYDALLOS PIRAEUS-GREECE