1. 如图,在$\triangle ABC$中,$D为AC$边上一点,$\angle DBC = 20^{\circ}$,$AD = DB = BC$。求$\angle A$的度数。

解:设$\angle A = x$.
$\because AD = BD$,
$\therefore \angle ABD = \angle A = x$.
$\because DB = BC$,
$\therefore \angle C = \angle BDC = \angle A + \angle ABD = 2x$.
在$\triangle BDC$中,
$\angle DBC + \angle C + \angle BDC = 180^{\circ}$,
$\therefore 2x + 2x + 20^{\circ} = 180^{\circ}$,
解得$x = 40^{\circ}$,
即$\angle A = 40^{\circ}$.

2. 如图,在$\triangle ABC$中,$\angle BAC = 100^{\circ}$,点$D$,$E分别在边BC$,$AC$上,且$AB = AD = DE = EC$。求$\angle C$的度数。

解:设$\angle C = x$.
$\because AB = AD = DE = EC$,
$\therefore \angle EDC = \angle C = x$,
$\angle DAE = \angle AED = \angle EDC + \angle C = 2x$,
$\angle B = \angle ADB = \angle DAE + \angle C = 3x$.
在$\triangle ABC$中,
$\angle B + \angle C + \angle BAC = 180^{\circ}$,
$\therefore 3x + x + 100^{\circ} = 180^{\circ}$,
解得$x = 20^{\circ}$,
$\therefore \angle C = 20^{\circ}$.

3. 如图,在$\triangle ABC$中,$AB = AC$,$D$,$E分别在AC$,$AB$上,且$BD = BC$,$BE = DE = AD$。求$\angle C$的度数。

解:$\because BE = DE = AD$,
$\therefore$设$\angle EBD = \angle EDB = x$,
则$\angle A = \angle AED = 2x$,
$\therefore \angle BDC = 3x$.
$\because AB = AC$,$BD = BC$,
$\therefore \angle BDC = 3x = \angle C = \angle ABC$,
$\therefore$在$\triangle ABC$中,
$2x + 3x + 3x = 180^{\circ}$,
解得$x = 22.5^{\circ}$,
$\therefore \angle C = 67.5^{\circ}$.

4. 如图,$C为\triangle ABE的边BE$上一点,且$AB = AC$,$AB的垂直平分线交AC于点D$,且$BD = BC$,$CE = CD$。
(1)求$\angle BAC$的度数；
(2)求$\angle CAE$的度数。

解:(1)设$\angle BAC = x = \angle ABD$,
则$\angle BDC = 2x = \angle BCD = \angle ABC$,
$\therefore 5x = 180^{\circ}$,
解得$x = 36^{\circ}$,
即$\angle BAC = 36^{\circ}$;
(2)连接$DE$,
易求$\angle CDE = \angle DEC = \frac{1}{2}\angle BCD = 36^{\circ}$,
$\therefore DE = BD = AD$,
$\therefore \angle CAE = \frac{1}{2}\angle CDE = 18^{\circ}$.