证明：（d-a)/(d-g)=[(e+f+1)-(b+c+1)]/[(e+f+1)-(h+i+1)]
=[(e-b)+(f-c)]/[(e-h)+(f-i)]
由合比定理（a/b=c/d==>a/b=(a+c)/(b+d)）得
（d-a)/(d-g)=[（d-a)+(e-b)+(f-c)]/[（d-g）+(e-h)+(f-i)]
设上式值为m,
即（d-a)/(d-g)=[（d-a)+(e-b)+(f-c)]/[（d-g）+(e-h)+(f-i)]=m
同理：(f=d-e-1;c=a-b-1;i=g-h-1)
（f-c)/(f-i)=[(d-e-1)-(a-b-1)]/[(d-e-1)-(g-h-1)]
=[(d-a)+(e-b)]/[(d-g)+(e-h)]
再利用合比定理
（f-c)/(f-i)=[（d-a)+(e-b)+(f-c)]/[（d-g）+(e-h)+(f-i)]=m
同理可证（e-b)/(e-h）=[（d-a)+(e-b)+(f-c)]/[（d-g）+(e-h)+(f-i)]=m
故：(d-a)/(d-g)=(e-b)/(e-h)=(f-c)/(f-i)