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.
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Posted by Antonio Gutierrez at 9:56 AM
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4 comments:

  1. c .t . e. oJanuary 25, 2011 at 12:23 PM

    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

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  2. c .t . e. oJanuary 25, 2011 at 1:23 PM

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

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  3. sourav mishraFebruary 10, 2011 at 4:12 AM

    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.

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  4. Sumith PeirisJuly 14, 2015 at 3:12 AM

    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
    Moratuwa
    Sri Lanka

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