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

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

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## Thursday, October 30, 2014

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

Labels:
circle,
circumcircle,
concyclic,
cyclic quadrilateral,
infographic,
quadrilateral,
triangle

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∠EHF = 180° − ∠A

ReplyDelete∠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.

From circle JHFA, <JHF=180-<CAD

ReplyDeleteFrom 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

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

ReplyDelete<HKC ( = <HFD = <AEH ) = <BGH implies K, H, G, C are concyclic

Problem 1052

ReplyDeleteIs <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