<image>As shown in the figure $$1$$, in the square $$ABCD$$, $$P$$ is a point on the diagonal $$BD$$, point $$E$$ is on the extension of $$AD$$, and $$PA=PE$$. $$PE$$ intersects $$CD$$ at $$F$$.  
 $$(1)$$ Prove: $$PC=PE$$; $$(2)$$ Find the measure of $$\angle CPE$$; $$(3)$$ As shown in figure $$2$$, change the square $$ABCD$$ to a rhombus $$ABCD$$, keeping the other conditions unchanged. When $$\angle ABC=120{}^\circ $$, connect $$CE$$. Investigate the relationship between segment $$AP$$ and segment $$CE$$, and explain the reasoning.