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Geometry ProblemClick the figure below to see the complete problem 573.Zoom at: Geometry Problem 573
name K arc BC meet EH, L arc CD meet FHin ▲ EFHH = C - ( HED + HFB )= C - ( CD/2 - CK/2 + MB/2 - CL/2 )H = C - ( C/2 - A/2 ) = C/2 + A/2H = 90
a correctionCD/2 is GD/2 ( third row )
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.
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 PeirisMoratuwaSri Lanka