Question

p为等边三角形ABC内一点，求证:PA+PB+PC<2AB

Answer 1

证明:分别以AP,AC为边长作等边⊿APD,等边⊿ACE.连接DE,BE.

延长BF交AE于M,延长PD交AE于F.(如图)

∠PAD=∠CAE=60°,则∠PAC=∠DAE;

又PA=AD;AC=AE.得⊿PAC≌ΔDAE(SAS),PC=DE.

PB+PM<AB+AM;------------------------(1)

PD+DF<PM+PM+MF;------------------(2)

DE<DF+FE.------------------------------(3)

(1)+(2)+(3),得:(PB+PD+DE)+(PM+DF)<(AB+AM+MF+FE)+(PM+DF);

即:(PB+PA+PC)+(PM+DF)<(AB+AE)+(PM+DF);

∴PB+PA+PC<AB+AE;

故:PB+PA+PC<2AB.