Proposed Problem

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Complete Problem 294

Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Thursday, May 28, 2009

### Problem 294: Right triangle, Circumcenter, Excenter, Hypotenuse, Perpendicular

Labels:
circumcenter,
circumcircle,
excenter,
perpendicular,
right triangle

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name P circle D meet AD, G meet BC, T meet AC

ReplyDeleteang PDG = x ( ang A = 2x, ang APB = GPD, G = 90°,B=90°)

ang PDT = 90 - x ( from tr ADT)

=> GDT = 90 - x - x = 90 - 2x

=> GDC = 45 - x ( DC bisector of GDT )

=> GCD = 90- ( 45 - x ) = 45 + x

ECD = 45 + x - x ( GCE = x, at arc BE as ang A/2)

ECD = 45

EC = ED = BE (1)

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▲FBE ~ ▲AEC (ang FBE = 90 - 2x + x =90-x, ACE = 90-x)

BF/BE = BE/AC

BF ∙ AC = BE² ( from (1) ) =>

BF ∙ AC = ED²

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Problem 294

ReplyDeleteIs BE=CE=DE. Fetch EK perpendicular to AC ( point K belongs to the straight AC ).Then OE is bisector <COB so EK=EF. Therefore triangle EFB= triangle EKC then FB=KC. But the rectangular triangle AEC apply 〖CE〗^2=CK.AC or 〖DE〗^2=BF.AC.