在三角形ABC中,向量AB=a,AC=b,AD=xa,AE=yb（0

还没学定点分点公式就这样:
DP=λPC ==> DA+AP=λ(PA+AC) ==> (1+λ)AP=AD+λAC
所以有AP=(xa+λb)/(1+λ)
下面就一样了.相当于推导一次公式.
设DP=λPC,BP=μPE,由定比分点公式得
AP=(xa+λb)/(1+λ)=(a+μyb)/(1+μ)
因为a,b不共线,所以
x/(1+λ)=1/(1+μ),
λ/(1+λ)=μy/(1+μ),
解得
λ=(xy-y)/(y-1),μ=(x-1)/(xy-x)
代入得
AP=[(xy-x)a+(xy-y)b]/(xy-1)