P是等边三角形ABC外的一点,∠APB=60°,求证;PA=PB+PC
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P是等边三角形ABC外的一点,∠APB=60°,求证;PA=PB+PC
P是等边三角形ABC外的一点,∠APB=60°,求证;PA=PB+PC
P是等边三角形ABC外的一点,∠APB=60°,求证;PA=PB+PC
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
∴ PB+PC = PD+PC = CD
∵ ∠CBD = ∠CBP+∠PBD= ∠CBP+ 60°
∠ABP = ∠CBP+ ∠ABC= ∠CBP+ 60°
∴ ∠ABP = ∠CBD
又∠APB =∠CDB = 60°
AB = CB
∴ ΔAPB ≌ ΔCDB
∴ AP = CD
又 PB+PC = CD
∴ PA = PB+PC
证毕