一个有趣的几何定理

D、E分别在AB、AC上，CD与BE交于F：如果AD+DF=AE+EF，且BF=CF，那么AB=AC。

设∠BAF=a，∠CAF=b，∠ABF=c，∠ACF=d，AF=1，可以求出各线段的长度：

AD=FullSimplify[(Sin[(b+d)]*(Sin[(a+b+d)])^((-1))),Refine[0Pi/2&&0Pi/2&&0Pi/2&&0Pi/2&&0Pi&&0Pi]]

DF=FullSimplify[((1+((Sin[(b+d)])^(2)*(Sin[(a+b+d)])^((-2)))+(Cos[a]*Sin[(b+d)]*(Sin[(a+b+d)])^((-1))*(-2))))^(1/2),Refine[0Pi/2&&0Pi/2&&0Pi/2&&0Pi/2&&0Pi&&0Pi]]

AE=FullSimplify[(Sin[(a+c)]*(Sin[(a+b+c)])^((-1))),Refine[0Pi/2&&0Pi/2&&0Pi/2&&0Pi/2&&0Pi&&0Pi]]

EF=FullSimplify[((1+((Sin[(a+c)])^(2)*(Sin[(a+b+c)])^((-2)))+(Cos[b]*Sin[(a+c)]*(Sin[(a+b+c)])^((-1))*(-2))))^(1/2),Refine[0Pi/2&&0Pi/2&&0Pi/2&&0Pi/2&&0Pi&&0Pi]]

AB=FullSimplify[Abs[(Sin[(a+c)]*(Sin[c])^((-1)))],Refine[0Pi/2&&0Pi/2&&0Pi/2&&0Pi/2&&0Pi&&0Pi]]

BF=FullSimplify[((1+((Sin[(a+c)])^(2)*(Sin[c])^((-2)))+(Cos[a]*Sin[(a+c)]*(Sin[c])^((-1))*(-2))))^(1/2),Refine[0Pi/2&&0Pi/2&&0Pi/2&&0Pi/2&&0Pi&&0Pi]]

AC=FullSimplify[Abs[(Sin[(b+d)]*(Sin[d])^((-1)))],Refine[0Pi/2&&0Pi/2&&0Pi/2&&0Pi/2&&0Pi&&0Pi]]

CF=FullSimplify[((1+((Sin[(b+d)])^(2)*(Sin[d])^((-2)))+(Cos[b]*Sin[(b+d)]*(Sin[d])^((-1))*(-2))))^(1/2),Refine[0Pi/2&&0Pi/2&&0Pi/2&&0Pi/2&&0Pi&&0Pi]]

AD+DF=AE+EF等价于：

FullSimplify[AD+DF-(AE+EF),Refine[0Pi/2&&0Pi/2&&0Pi/2&&0Pi/2&&0Pi&&0Pi]]//TrigFactor//FullSimplify

AB+BF=AC+CF等价于：

FullSimplify[AB+BF-(AC+CF),Refine[0Pi/2&&0Pi/2&&0Pi/2&&0Pi/2&&0Pi&&0Pi]]//TrigFactor