证明:(1)延长BP至E,使PE=PC,连接CE, ∵∠1=∠2=60°,∠3=∠4=60°, ∴∠CPE=60°, ∴△PCE是等边三角形, ∴CE=PC,∠E=∠3=60°, 又∵∠EBC=∠PAC, ∴△BEC≌△APC, ∴PA=BE=PB+PC; | ![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105193437-62351.gif) |
(2)证明:过点B作BE⊥PB交PA于E, ∵∠1+∠2=∠2+∠3=90°, ∴∠1=∠3, 又∵∠APB=45°, ∴BP=BE, ∴PE=![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105193437-27855.gif) 又∵AB=BC, ∴△ABE≌△CBP, ∴PC=AE, ∴PA=AE+PE=PC+ ; | ![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105193437-51485.gif) |
(3)PA=PC+ , 证明:在AP上截取AQ=PC,连接BQ, ∵∠BAP=∠BCP,AB=BC, ∴△ABQ≌△CBP, ∴BQ=BP, 又∵∠APB=30°, ∴PQ=![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105193438-40973.gif) ∴PA=PQ+AQ= +PC。 | ![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105193438-38833.gif) |