Geometry Problem

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## Tuesday, January 25, 2011

### Problem 573: Cyclic quadrilateral, Angle bisector, Perpendicular

Labels:
angle bisector,
circle,
cyclic quadrilateral,
perpendicular

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name K arc BC meet EH, L arc CD meet FH

ReplyDeletein ▲ 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

a correction

ReplyDeleteCD/2 is GD/2 ( third row )

let EG intersect BC at K and AD at L respectively.

ReplyDeleteangle 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.

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

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka