定理已知p_1(x_1,y_1),p_2(x_2,y_2)两点及直线l:mx+ny+r=0,若l交有向线段p_1p_2或其延长线于点p(x,y),则λ=p_1p/pp_2=-(mx_1+ny_1+r)/(mx_2+ny_2+r)。λ≠-1,当λ>0时,p内分p_1p_2;当λ<0时,p外分p_1p_2。证点p(x,p)的坐标为 x=(x_1+λx_2)/(1+λ),y=(y_1+λy_2)/(1+λ)。又p在l上,故m(x_1+λx_2)/(1+λ)+n(y_1+λy_2)/(1+λ)+r==0,即 p_1p/pp_2=λ=-(mx_1+ny_1+r)/(mx_2+ny_2+r)。下面用此定理证明一类几何题。 The theorem knows p_1(x_1,y_1),p_2(x_2,y_2) two points and a straight line l:mx+ny+r=0, if l intersects a directed segment p_1p_2 or extends it at a point p(x,y), Then λ=p_1p/pp_2=-(mx_1+ny_1+r)/(mx_2+ny_2+r). λ≠-1, when λ>0, p is divided into p_1p_2; when λ<0, p is divided into p_1p_2. The coordinates of the proof point p(x,p) are x=(x_1+λx_2)/(1+λ), and y=(y_1+λy_2)/(1+λ). Again p is on l, so m(x_1+λx_2)/(1+λ)+n(y_1+λy_2)/(1+λ)+r==0, ie p_1p/pp_2=λ=-(mx_1+ny_1 +r)/(mx_2+ny_2+r). The following theorem is used to prove a class of geometric problems.