电子课本网 第22页

第22页

信息发布者:
B
C
D
26
9
84°
证明:
$\because \angle BAE = \angle CAD,$
$\therefore \angle BAE + \angle CAE = \angle CAD + \angle CAE,$即$\angle BAC = \angle EAD.$
在$\triangle ABC$和$\triangle AED$中,
$\begin{cases}AB = AE,\\\angle BAC = \angle EAD,\\AC = AD,\end{cases}$
$\therefore \triangle ABC\cong\triangle AED$(SAS)
证明:
$\because AD = AE,$$BD = CE,$
$\therefore AD + BD = AE + CE,$即$AB = AC.$
在$\triangle ABE$和$\triangle ACD$中,
$\begin{cases}AB = AC,\\\angle A = \angle A,\\AE = AD,\end{cases}$
$\therefore \triangle ABE\cong\triangle ACD$(SAS).
$\therefore \angle B = \angle C$