题目内容
解:把△APC绕A逆时针旋转60°得到△AP'C',如图∴∠CAC'=∠PAP'=60°,AC=AC',AP=AP',PC=P'C,∴△APP'为等边三角形,∴PP'=AP,∵∠BAC=120°,∴∠BAC'=120°+60°=180°,即B,A,C'共线,∴BC'<BP+PP'+P'C,即AB+AC<AP+BP+CP.