延长ep到h,使得ph=pf,连接hd延长交ab于g∵eh⊥bc,ph=pf∴△cfh是等腰三角形,∠bch=∠bca∵ab=ac∴∠bch=∠bca=∠cbaad⊥bc,bd=dc∴△bdg与△cdh全等,dg=dh看△geh,ad//eh,dg=dh所以ad=1/2eh=1/2(pe+ph)=1/2(pe+pf)即...