(1)证明:∵MN∥BC,CE平分∠ACB,CF平分∠ACD, ∴∠BCE=∠ACE=∠OEC,∠OCF=∠FCD=∠OFC, ∴OE=OC,OC=OF, ∴OE=OF.
(2)当O运动到AC中点时,四边形AECF是矩形, ∵AO=CO,OE=OF, ∴四边形AECF是平行四边形, ∵∠ECA+∠ACF=∠BCD, ∴∠ECF=90°, ∴四边形AECF是矩形.
(3)当四边形AECF是正方形时,AO⊥EF, ∵BC∥EF, ∴AC⊥BC,AC=AE, ∵=, ∴BC=AE, ∴tanB===, ∴∠B=60°. |