如图:
分别取$ CD,DB,DC $中点$ F,N,O $,连接$ OF,ON,AO $,有\[ ON=DF=2OF \]取$ CM=BN $,延长$ AC $至$ P $使$ CP=CM $,有\[ \triangle MOP\cong \triangle DBF\sim \triangle FOE \]所以$ FNCO $四点共园和$ ECPO $四点共园,有\[ \angle FCM=\angle FOM \\\angle CPE=\angle COE\]又\[ \angle EMP+\angle MPE=90\du =\angle AOE+\angle EOC \]有\[ \angle EMP=\angle AOE \]故$ AOEM $四点共园,因此\[ \angle MAE=\angle MOE \]得\[ \angle AED=\angle ACE+\angle CAE=\angle FOM+\angle MOE=\angle FOE=\angle DEO=\angle DBF=\dfrac{1}{2}\angle ADE \] |