三角形动点和将军饮马

证明：延长AC至L，使得CL=BC
(ecause angle ACB=60° herefore angle BCL=120°)
(ecause BC=CL, herefore angle CLB =angle CBC=30°)
(ecause angle EDB=120°，ED=DB)
( herefore angle DEB=angle DBE=30°)
(ecause angle EDB=120°，ED=DB)
( herefore angle DEB = angle DBE =30°)
( herefore riangle EDB= riangle LCB)
( herefore frac{DB}{BC}=frac{ED}{CL}=frac{EB}{LB})
(ecause angle DBE=angle CBL =30°)
( herefore angle DBE + angle EBD=angle CBL+angle EBC)
(即：angle DBC=angle EBL)
( herefore riangle DBC sim riangle EBL)
( herefore angle ELB=angle DCB=60°)
(ecause angle CL B=30°， herefor angle ELD=30°)
( herefore 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)
( herefore 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