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

Labels:
concyclic,
cyclic quadrilateral,
perpendicular,
side

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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