* pd⊥ab,cf⊥⊥ab

∴ Police station //CF

P is PG//AB, vertical foot is g, spanning CF and q.

∴ CQ⊥PG, quadrilateral FQPD is a rectangle.

∴ PD=FQ

∫△ABC is an isosceles triangle, while PE⊥AC

∴ PE=CQ

∴ PD+PE=FQ+CQ=CF

2. If point P is on the extension line of BC, then the pure relationship among PE, PD and CF is:

PD=CF+PE

Proof: Point P is on the extension line of BC.

Do CQ⊥DP to C, the vertical foot is Q,

And ∵PD⊥AB,CF⊥AB.

Quadrilateral DQCF is a rectangle.

∴DQ=CF，

△ ABC is an isosceles triangle.

∴PQ=PE

∴PD=PQ+DQ=CF+PE