Question

如图,P是△ABC的外角∠EAC的平分线AD上的一点(不与A重合),求证PB+PC>AB+AC

Answer 1

延长BA至E,使AE＝AC；再延长BP至F,使PF＝PC.

∵∠PAC＝∠PAE、AC＝AE,∴PA是CE的垂直平分线,∴PC＝PE,又PC＝PF,∴PE＝PF,

∴∠BFE＝∠PEF.

显然有：∠BEF＞∠PEF,∴∠BEF＞∠BFE,∴BF＞BE,∴PB＋PF＞AB＋AE,

∴PB＋PC＞AB＋AC.