假设AD=13,AE=52,∠ACB为α,BG⊥AF
DF/FC=x
x =DG/EC=13*sin2α / (52/tanα)=(1/2)(sinα)^2
易证AC为∠A的外角平分线
DC/CF=AD/AF
x+1=sin∠AFD/sin∠ADF=sin（∠ADF+2α）/(12/13)
=(5sin2α+12cos2α)/12
化简tanα=1/3
Sinα=1/√10
x=1/20

横过来，梅涅劳斯定理一带就好了