Tuesday, January 25, 2011

Problem 573: Cyclic quadrilateral, Angle bisector, Perpendicular

Geometry Problem
Click the figure below to see the complete problem 573.

 Problem 573: Cyclic quadrilateral, Angle bisector, Perpendicular.
Zoom at: Geometry Problem 573


  1. name K arc BC meet EH, L arc CD meet FH
    in ▲ EFH
    H = C - ( HED + HFB )= C - ( CD/2 - CK/2 + MB/2 - CL/2 )
    H = C - ( C/2 - A/2 ) = C/2 + A/2
    H = 90

  2. a correction
    CD/2 is GD/2 ( third row )

  3. let EG intersect BC at K and AD at L respectively.
    angle HLF = angle A + angle BEL (by exterior angle property)
    angle HKF = angle KCE + angle KEC.
    but angle KEC = angle BEL since EL bisects angle BEC.
    and angle KCE = angleA because in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
    these imply that angle HLF = angle HKF.
    also HF is common to the triangles HLF and HKF.
    and angle HFL = angle HFK because FM bisects angle AFB.
    so by AAS criterion the triangles HLF and HKF are congruent. so it implies that angle KHF = angle LHF. but they constitute a linear pair so angle LHF = angle KHF = 90 degrees.
    therefore EG is perpendicular to FM.
    Q. E. D.

  4. Let G meet BC and AD at X and Y respectively. Then < X and < Y are both equal to <A + E/2 since external < C = A and so Tr. FXY is isoceles and the result follows

    Sumith Peiris
    Sri Lanka