Proposed Problem
Click the figure below to see the complete problem 335 about Cyclic Quadrilateral, Perpendicular to sides, Concyclic points.
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Complete Problem 335
Level: High School, SAT Prep, College geometry
Wednesday, August 5, 2009
Problem 335. Cyclic Quadrilateral, Perpendicular to sides, Concyclic points
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Many triangles are similar. Especially
ReplyDeletePAE ~ PCG, PBF ~ PDH.
Corresponding sides are in proportion,
PA:PC = AE:CG =PE:PG,
PB:PD = BF:DH = PF:PH.
Product of extremes and the product of means are equal,
PA CG = PC AE, PB DH = PD BF
PA PG = PC PE, PB PH = PD PF.
Multiplying the first two equations and the last two,
PA CG PB DH = PC AD PD BF
PA PG PB PH = PC PE PD PF.
From the power theorem on circles, PA PB = PC PD, so by canceling,
CG DH = AD BF
PG PH = PE PF.
By the converse of power theorem, four points E, F, G, and H are cyclic.
Problem 335
ReplyDeleteSi AB CD intersect to K.Triangles KAE,KDH are similar so AE/DH= KA/KD,triangles KAD,KCB are similar so KA/KD=KC/KB and triangles KCF, KBF are similar so KC/KB=CG/BF then AE/DH=CG/BF.
Then AE.BF=DH.CG.Is AHDE , CGBF concyclic so <FGB=FCB=<DAH=<DEH. Therefore E,H,E,F are concyclic.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE