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
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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.
Right triangle EFB similar to AEC and also BE=EC
ReplyDeleteso BF/BE = EC/AC
=> EC^2=BF.AC ---------------(1)
CED is isosceles right triangle => EC=DE
Thus DE^2=BF.AC