See complete Problem 38

Right triangle, altitude, incenters, angles. Level: High School, SAT Prep, College geometry

## Monday, May 19, 2008

### Elearn Geometry Problem 38

Labels:
altitude,
angle,
incenter,
right triangle

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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.