记直线AC和BE交点为P,PO交劣弧AB和线段AB分别为点为M,N
再记∠APO=∠BPO=α.则FGC截△APM,FHE截△PBM,由梅涅劳斯定理得
∴(AG/GM)(MF/FP)(PC/CA)=1,(BH/HM)(MF/FP)(PE/EB)=1
将两式相乘得(AG/GM)(BH/HM)(MF/FP)²(PC/CA)(PE/EB)=1
又∵FM=AM*AM/MN,FP=PA*PA/PN,且AN平分∠PAM,∴PN/NM=PA/AM
∴MF/FP=AM/PA=sinα.再记PC=x,PE=y,CE=z,
则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-2xycos∠CAE)=2/(1-cos2α)=1/sin²α
∴(MF/FP)²(PC/CA)(PE/EB)=sin²α/sin²α=1,即(AG/GM)(BH/HM)=1
∴AG/GM=HM/BH,又AG+GM=AM=BM=BH+HM,∴AG=HM,GM=BH
即GH=GM+MH=GM+AG=AM=AB/2