First, when point P coincides with point B or point C,

B(P)E or C(P)E is the height of an isosceles triangle,

Because, AB=AC,

Therefore, b (p) e = c (p) e = 2s △ ABC/ab = h.

Second, when point p does not coincide with point b or point c,

As auxiliary lines: the intersection point P is PE, and the vertical AB is E,

Make PF perpendicular to AC and f, and connect AP.

Therefore, PE=2S△APB/AB, PF=2S△APC/AC.

That is, PE+PF=2S△APB/AB+2S△APC/AC.

Because, AB=AC

Therefore, PE+PF=2S△APB/AB+2S△APC/AB.

=2S△ABC/AB

=h

So PE+PF is equal to a fixed value.

(2)? Conclusion: | PE-PF | is equal to a fixed value.

Prove:

As an auxiliary line: extend BC to p and BA to n,

Extend AC to m. the intersection p is PE, the vertical BN is e,

Make PF perpendicular to AM and f and connect AP.

Because PE is as high as △PAB and PF△PAC,

Therefore, PE=2S△PAB/AB and PF=2△PAC/AC.

Because, AB=AC

So | PE-PF |

=|2S△PAB/AB-2△PAC/AC|

=|2S△ABC/AB|

=h

Therefore | PE-PF | is equal to a fixed value.