Question

设P是正三角形ABC外接圆的劣弧BC上任意一点,求证：PB+PC=PA,PB*PC=PA^2-PB^2

Answer 1

证明：延长PC至D点,使得PA=PD,连接AD.

∵∠DPA=∠CBA=60°,∴⊿PAD是等边三角形,

∴DA=PA

∵AB=AC,PA=AD,∠BAP=∠CAB-∠PAC=∠DAP-∠PAC,

∴⊿APB≌⊿ACD∴BP=CD

∴PA=PC+CD=PC+PB,即PA=PB+PC