Monday, May 19, 2008

Elearn Geometry Problem 38



See complete Problem 38
Right triangle, altitude, incenters, angles. Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:08 PM
Labels: , , ,

2 comments:

  1. c .t . e. oFebruary 6, 2010 at 3:07 PM

    ang B = α + α + A/2 + A/2 (see P37)

    2α + A = 90°

    α = 45 - A/2
    ------------------------------------------

    EBG = CBE - CBG = 45° - A/2 ( BE bisector)

    EBG = α
    ------------------------------------------
    EBG = E'FG (perpendicular sides , see P37)

    E'FG = 90°
    ------------------------------------------
    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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  2. bae deok rak(배덕락)October 6, 2010 at 5:35 PM

    In Problem 38, we get
    (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.

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