根据三角形面积关系得出线段(底、高)关系,是一种较好的解题方法. 例1 如图1,△ABC中,AB=AC,BD是高,P为BC延长线上一点,PE⊥AB,PF⊥AC,垂足分别为E、F.求证:PE=BD+PF. 分析:证明线段和差关系的常规思路是截长或补短,可利用全等实现线段的转移;而本题则可由高想 According to the relationship of the area of ​​the triangle, the line segment (bottom, height) relationship is a good solution to the problem. Example 1 As shown in Figure 1, △ABC, AB=AC, BD is high, and P is a point on the BC extension line. PE⊥AB, PF⊥AC, foot pedals are respectively E and F. Proof: PE=BD+PF. Analysis: The conventional idea of ​​verifying the relationship between line segment and difference is to shorten the length or make up the short line. And this question can be made by Gao Xiang