Thursday, July 28, 2016

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

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.

Click the figure below to view more details of problem 1239.


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

4 comments:

  1. ∠BFG=∠BFC=∠BAD=∠BED
    ∴B E F G concyclic
    ∠EBH=180°-∠EBF=180°-∠EAC=∠EAD
    ∴ED=EH

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  2. nice solution. Thank you very much.

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  3. Problem 1239
    Is <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

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  4. Let Ct be the tangent to circle (O) on C.
    Consider 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).

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