Thursday, May 28, 2009

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

Proposed Problem
Click the figure below to see the complete problem 294.

 Problem 294: Right triangle, Circumcenter, Excenter, Hypotenuse.
See also:
Complete Problem 294
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

2 comments:

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

    ang 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)
    -------------------------------------------
    ▲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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  2. Problem 294
    Is 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.

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