【P是等边三角形ABC外的一点,∠APB=60°,求证;PA=PB+PC】

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证明：       过点B作 BD//AP,交CP延长线于D；设AP,BC交点为 Q
    等边ΔABC ==> AB=BC=CA；∠ABC=∠ACB=60°
       ∵  ∠APB = 60° 
       ∴  ∠APB =∠ACB； 
        又  ∠AQC =∠BQP     
       ∴ ΔADC ∽ ΔBQP  ==> AQ/BQ = CQ/PQ
                                      ==> AQ/CQ =BQ/PQ
        又 ∠AQB =∠CQP
       ∴ ΔAQB ∽ ΔCQP  ==> ∠ABC =∠APC=60°
        
        
BD//AP ==> ∠PBD = ∠APB=60° ；∠BDP = ∠APC=60° ；        
    ==>等边ΔBDP ==> PB=PD