(1)∵∠PAB+∠PBA=180°-∠APB=60°,
∠PBC+∠PBA=∠ABC=60°,
∴∠PAB=∠PBC,
又∵∠APB=∠BPC=120°,
∴△ABP∽△BCP,
∴
=,
∴PB
2=PA?PC=12,
∴PB=2
;
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b8389b504fc2d562d12eca6fe41190ef76c66c09?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
(2)证明:在BB'上取点P,使∠BPC=120°.连接AP,再在PB'上截取PE=PC,连接CE.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/0eb30f2442a7d933706ffff1ae4bd11372f00168?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∠BPC=120°,
∴∠EPC=60°,
∴△PCE为正三角形,
∴PC=CE,∠PCE=60°,∠CEB'=120°.
∵△ACB'为正三角形,
∴AC=B′C,∠ACB'=60°,
∴∠PCA+∠ACE=∠ACE+∠ECB′=60°,
∴∠PCA=∠ECB′,
∴△ACP≌△B′CE,
∴∠APC=∠B′EC=120°,PA=EB′,
∴∠APB=∠APC=∠BPC=120°,
∴P为△ABC的费马点.
∴BB'过△ABC的费马点P,且BB'=EB'+PB+PE=PA+PB+PC.