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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,
∵∠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.