See complete Problem 38
Right triangle, altitude, incenters, angles. Level: High School, SAT Prep, College geometry
Monday, May 19, 2008
Elearn Geometry Problem 38
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See complete Problem 38
Right triangle, altitude, incenters, angles. Level: High School, SAT Prep, College geometry
ang B = α + α + A/2 + A/2 (see P37)
ReplyDelete2α + A = 90°
α = 45 - A/2
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EBG = CBE - CBG = 45° - A/2 ( BE bisector)
EBG = α
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EBG = E'FG (perpendicular sides , see P37)
E'FG = 90°
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F', F" tg points for F, G
F'FE = A/2 ( AD//FF')
=>F'FF" = A/2 - α => ( E'FG = α )
DFG = 45° + A/2 - α ( DF bisector )
DFG = 45° - α + A/2
DFG = A/2 + A/2 ( 45 - α = A/2)
DFG = A
▲FDG = A + 90° + DGF & A + 90° = A + B
DGF = C = 2α
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In Problem 38, we get
ReplyDelete(1) ang(EBG)=ang(EBC)-ang(GBC)=45-((90-2α)/2)=α.
(2) Since BPQF is a cyclic quadrilateral,
ang(EFG)=ang(EFQ)=ang(QBG)=ang(EBG)=α,
where P and Q are foots of the altitudes FP
and BQ, respectively.
(3) ang(DGF)=(ang(GDC)+ang(GCD))-ang(FGE)
=(45+α)-(90-α-45)=2α.