Question

如图,四边形ABCD中,AB=BC,角ABC=60度,若P为四边形ABCD内一点,且角APD=120度,

Answer 1

延长DP到点P'使得AP=AP'连接BP′,AC∵APD＝120°,∴∠APP'=60,AP=AP',∴△APP'是等边三角形.∴P'P=AP

同理易见△ABC也是等边三角形,

∵AB=BC, AP=AP',∠BAC=∠P'AP,∠BAP'=∠CAP

∴△BAP'≌△CAP

∴BP'=PC.

∴PA+PD+PC=PP'+PD+BP'

而两点间直线距离最短,有PP'+PD+BP'≥BD

即PA+PD+PC>≥BD