Question

十分钟后就下了

Answer 1

当O运动到AC中点时,四边形AECF是矩形.

∵CE平分∠ACB,∴∠BCE＝∠OCE,

∵MN∥BC,∴∠OEC＝∠BCE

∴∠OEC＝∠OCE,

∴OE＝OC,

同理可证OF＝OC,

则OE＝OF,

又由AO＝OC,

则四边形AECF为平行四边形,

∵OE、OF分别平分∠ACB、∠ACD

∴∠ACE+∠ACF

＝½∠ACB+½∠ACD

=½(∠ACB+∠ACD)

=½×180°

=90°

即∠ECF＝90°

∴平行四边形AECF是矩形.