It is proved that if the intersection of straight line AC and BE is p, then POF*** line, PO intersecting arc AB and line segment AB are points M respectively, and then n is ∠APO=∠BPO=α. Then FGC intercepts △APM, FHE intercepts △PBM, and by Menelaus theorem ∴ (Ag/GM) (MF/FP) (PC/CA) =1,(BH/HM)(MF/FP)(PE/EB)= 1 If PC=x, PE=y and CE=z, then PA+PB=PC+CD+DE+PE=x+y+z, ∴AC=PA-PC=(z+y-x)/2, EB =(z+x-y)/2 ∴(pc/ca)(pe/eb)=[2x/(z+y-x)][2y/(z+x-y)]=4xy/(z2-x2-y2+2xy)= 4xy/(2xy-2 cycos∠CAE)= 2/( 1-cos 2α)= 1/sin 2α∴(mf/fp)2(pc/ Ca) (PE/EB) = sin2α/sin2α =1,that is, (ag/GM)(BH/hm)= 65438+ 1