Friday, November 5, 2010

Problem 537: Rectangle, Midpoints, Diagonal, Angle Bisector

Geometry Problem
Click the figure below to see the complete problem 537 about Rectangle, Midpoints, Diagonal, Angle Bisector.

Problem 537: Rectangle, Midpoints, Diagonal, Angle Bisector. Level: High School, SAT Prep, College Geometry
Go to Complete Problem 537


  1. Let a=BE=AF, b=AB=EF and c=AG
    Let K,M, Q are the projection of H over EF, FD and BA
    See link below for details

    1 Triangle HNF similar to Triangle HBG ( case AA)
    We have HK/HQ=b/(2.(b+c))
    HQ/2(b+c)=HK/b =a/(2c+b)
    So HK= a.b/(2c+b)

    2 Triangle FHM similar to triangle FAG ( case AA)
    We have HM/c=FM/a so HM=c/a*FM=b.c/(2c+b)

    3 Note that ratio HK/GP=EK/EP=b/(2c+b)
    Triangle EPG similar to triangle EKH ( case SAS)
    Corresponding angles GEF will congruence with angle KEH and EF is the angle bisector of angle GEH

    Peter Tran

  2. Let be N intersection of AB and HE (extended)
    FH/GF = OF/BG, EH/NE = OE/BN => BG = BN
    => EN = GE
    EH/EN = FH/GF => EH/EG = FH/GF
    EH/FH = EG/GF

  3. Let J intersection of CD and EH extended.
    Let I intersection of CD extended and GH.
    Tr. EFH similar to Tr. JIH.
    BH is median of EFH and JIH -> AG = ID = DJ.
    Then EJ = EG and Tr. EGJ is isosceles and EF is altitude -> EF is bisector of GEH.