解：?$(1)$?延長?$BP{至}E，$?使?$PE=PC，$?連結(jié)?$CE.$?
?$∵∠1=∠2=60°，$??$∠3=∠4=60°，$?
?$∴∠CPE=60°，$?
?$∴△PCE$?是等邊三角形，
?$∴CE=PC，$??$∠E=∠3=60°.$?
又?$∵∠EBC=∠PAC，$?
?$∴△BEC≌△APC，$?
?$∴PA=BE=PB+PE=PB+PC.$?
?$(2)$?過點?$B$?作?$BE⊥PB$?交?$PA$?于?$E.$?
?$∵∠1+∠2=∠2+∠3=90°，$?
?$∴∠1=∠3.$?
又易知?$∠APB=45°，$?
?$∴BP=BE，$?
?$∴PE=\sqrt{2}PB.$?
又?$∵AB=CB，$?
?$∴△ABE≌△CBP，$?
?$∴PC=AE，$?
?$∴PA=AE+PE=PC+\sqrt{2}PB.$?
?$(3)PA、$??$PB、$??$PC$?滿足關(guān)系式：?$PA=PC+\sqrt{3}PB.$?
理由如下：
在?$AP{上截取}AQ=PC，$?連結(jié)?$BQ，$?過點?$B$?作?$BM⊥PQ{于}M.$?
?$∵∠BAP=∠BCP，$??$AB=CB，$?
?$∴△ABQ≌△CBP，$?
?$∴BQ=BP.$?
又易知?$∠APB=30°，$?
?$∴PQ=\sqrt{3}PB，$?
?$∴PA=AQ+PQ=PC+\sqrt{3}PB.$?
