Thursday, October 30, 2014

Geometry Problem 1052: Triangle, Quadrilateral, Three Circumcircles, Circle, Concyclic Points, Cyclic Quadrilateral

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1052: Triangle, Quadrilateral, Three Circumcircles, Circle, Concyclic Points, Cyclic Quadrilateral.

Posted by Antonio Gutierrez at 9:08 AM
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4 comments:

  1. Jacob HA (EMK2000)October 30, 2014 at 6:19 PM

    ∠EHF = 180° − ∠A
    ∠EHG = 180° − ∠B
    ∠FHK = 180° − ∠D

    Thus
    ∠GHK = 180° − ∠C
    CGHK are concyclic.

    Also,
    ∠CJH
    = 180° − ∠AJH
    = 180° − ∠DFH
    = ∠DKH

    So, CJHK are concyclic.

    Hence, CGJHK are concyclic.

    ReplyDelete
  2. Ivan BazarovOctober 31, 2014 at 6:42 PM

    From circle JHFA, <JHF=180-<CAD
    From circle KHFD, <KHF=180-<CDA
    Therefore <JHK=360-(180-<CDA)-(180-<CAD)=<CDA+<CAD=180-<ACD, so CJHK is cyclic.
    <JHG=<EHG-<EHJ
    From circle BEHG, <EHG=180-<ABC
    From circle AEJH, <EHJ=<BAC
    Therefore <JHG=(180-<ABC)-(<BAC)=<BCA so GCHJ is cyclic.
    Cyclic quadrilaterals GCHJ and CJHK share triangle CJH, so GCKHJ is concylic

    ReplyDelete
  3. PravinNovember 7, 2014 at 12:14 AM

    <HKC ( = <HFD = <AEH ) = <AJH implies K, H, J, C are concyclic
    <HKC ( = <HFD = <AEH ) = <BGH implies K, H, G, C are concyclic

    ReplyDelete
  4. APOSTOLIS MANOLOUDISSeptember 9, 2016 at 4:10 AM

    Problem 1052
    Is <HKD=<EFA (FHKD=cyclic),<HFA=<HEB (AFHE=cyclic),<HEB=<HGC (BEHG=cyclic) so <HKD=<HGC then the GHKC is cyclic.But <HKC=HFD (HKDF= cyclic), <HFD=<HJA (AFHJ=cyclic) so <HKC=<HJA .Then JHKC is cyclic.Hence J,H,K,C and G are concyclic.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

    ReplyDelete

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