As shown in the figure, as an auxiliary line, the extension line of FO⊥CB is at point O.

Then ∠FBO=45 degrees, BO=FO.

From AP⊥PF, it is easy to know ∠CAP=∠OPF, so △ACP is similar to △POF, so we know that AC/CP=PO/FO, where PO=PB+BO=PB+FO, AC=BC=PB+PC, and we can know FO = by substitution. So AC=PO.

It is also known that △ACP is equal to △POF and AP=PF.

2)

Similarly, the proof method is the same as above. First, it is proved that △ACP is similar to △POF, and then it is proved that △ACP is equal to △POF.