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