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命题:设已知两点P_1(x,y_1)、P_2(x_2,y_2)的连线交直线l:Ax+By+C=0于点P(P_2不在直线l上) 求证:P_1P/PP_2=-(Ax_1+By_1+C)/(Ax_2+By_2+C) 证明:设P_1P/PP_2=λ,则点P坐标为 ((x_1+λx_2)/(1+λ),(y_1+λy_2)/(1+λ)) ∵点P在直线l上, ∴ A(x_1+λx_2)/(1+λ)+B(y_1+λy_2)/(1+λ)+C=0 解得λ=-(Ax_1+By_1+C)/(Ax_2+By_2+C) 所以P_1P/PP_2=-(Ax_1+By_1+C)/(Ax_2+By_2+C) (Ax_2+By_2+C≠0) 此命题在平几中用于证明比例线段问题,常能奏效。下面略举数例。例1.P为△ABC的边BC所对的中位线DE上任意一点,CP交AB于M,BP交AC于N,
Proposition: Let the connecting line of known two points P_1(x,y_1), P_2(x_2,y_2) be l:Ax+By+C=0 at point P (P_2 is not on the line l) Proof: P_1P/PP_2= - (Ax_1+By_1+C)/(Ax_2+By_2+C) Prove that if P_1P/PP_2=λ, the point P coordinate is ((x_1+λx_2)/(1+λ), (y_1+λy_2)/( 1+λ)) The point P is on the line l, ∴A(x_1+λx_2)/(1+λ)+B(y_1+λy_2)/(1+λ)+C=0 The solution is λ=-(Ax_1 +By_1+C)/(Ax_2+By_2+C) So P_1P/PP_2=-(Ax_1+By_1+C)/(Ax_2+By_2+C) (Ax_2+By_2+C≠0) This proposition is used in a few days. Prove the problem of proportional line segments, often work. Here are a few examples. Example 1. P is the edge of BC in △ABC at any point on the midline DE, CP crosses AB in M, BP crosses AC in N,