三角形动点和将军饮马

证明：延长AC至L，使得CL=BC
$\because \angle ACB=60° \therefore \angle BCL=120°$
$\because BC=CL,\therefore \angle CLB =\angle CBC=30°$
$\because \angle EDB=120°，ED=DB$
$\therefore \angle DEB=\angle DBE=30°$
$\because \angle EDB=120°，ED=DB$
$\therefore \angle DEB = \angle DBE =30°$
$\therefore \triangle EDB=\triangle LCB$
$\therefore \frac{DB}{BC}=\frac{ED}{CL}=\frac{EB}{LB}$
$\because \angle DBE=\angle CBL =30°$
$\therefore \angle DBE + \angle EBD=\angle CBL+\angle EBC$
$即：\angle DBC=\angle EBL$
$\therefore \triangle DBC \sim \triangle EBL$
$\therefore \angle ELB=\angle DCB=60°$
$\because \angle CL B=30°，\therefor \angle ELD=30°$
$\therefore E点运动轨迹为M到L的线段$
$M为E运动轨迹最左侧端点$
$\therefore \angle MAL=\angle MAB -\angle CAB$
$=120°-60°=60°$
$又因为 \angle MLC=30°，所以 \angle AME=90°$
所以AE+EC的最小值为AC',C’是C关于ML的对称点