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
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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
Let Ct be the tangent to circle (O) on C.
ReplyDeleteConsider lines DC and BC which include the intersections A and B of circles (P) and (Q).
Reim’s Theorem (see problem n°72 for the basic configuration), concludes to ED//Ct, thus GE//Ct.
Then, apply the reciprocal of Reim’s theorem to the configuration of F, G, C, E, B, enabling to conclude that F, G, E, B lie on a circle (O) intersecting circle (P) at F and B, intersected by lines (FGC) and (BEC).