Thursday, October 30, 2014

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

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

Click the figure below to enlarge the figure.

1. ∠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.

From circle KHFD, <KHF=180-<CDA
<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

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

4. 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