问题补充:
P是等边三角形ABC外的一点,∠APB=60°,求证;PA=PB+PC
答案:
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