19. （8 分）解不等式组$\begin{cases}5 + 3x\lt13,\frac{x + 2}{3} - \frac{x - 1}{2}\leq2,\end{cases}$并写出它的所有正整数解。
答案：$\begin{cases}5 + 3x ＜ 13①,\frac{x + 2}{3}-\frac{x - 1}{2}\leq2②.\end{cases}$
解不等式①，得$x ＜ \frac{8}{3}$。解不等式
②，得$x\geq - 5$。$\therefore$原不等式组的解集为$-5\leq x ＜ \frac{8}{3}$。$\therefore$其所有
正整数解为$1,2$
20. (10 分)如图，有下列条件：①$\angle1 + \angle2 = 180^{\circ}$；②$\angle3 = \angle A$；③$\angle B = \angle C$。从中选出两个作为题设，另一个作为结论可以组成 3 个命题。从中选择 1 个真命题，写出已知、求证，并证明。
如图，已知
①②
，求证：
③
(填序号)。
证明：
答案：答案不唯一，如①② ③ $\because\angle1 + \angle2 = 180^{\circ}$，$\therefore AD//EF$。$\therefore\angle3 = \angle D$。$\because\angle3 = \angle A$，$\therefore\angle A = \angle D$。$\therefore AB// CD$。
$\therefore\angle B = \angle C$
21. （12 分）如图，$AB// CD$，$\angle B = \angle D$，$AE$交$BC$的延长线于点$E$，交$CD$于点$F$。
（1）求证：$AD// BE$；
（2）若$\angle ACB = \angle CFE = 60^{\circ}$，$\angle BAC = 2\angle EAC$，求$\angle B$的度数。
答案：(1)$\because AB// CD$，$\therefore\angle B = \angle DCE$。$\because\angle B = \angle D$，
$\therefore\angle DCE = \angle D$。$\therefore AD// BE$ (2)$\because AB// CD$，$\angle ACB =\angle CFE = 60^{\circ}$，$\therefore\angle BAE = \angle CFE = 60^{\circ}$，$\angle BAC = \angle ACD$，
$\angle B = \angle DCE$。$\therefore\angle EAC + \angle BAC = 60^{\circ}$。$\because\angle BAC = 2\angle EAC$，
$\therefore\angle EAC = 20^{\circ}$。$\therefore\angle BAC = \angle ACD = 40^{\circ}$。$\because\angle ACB +\angle ACD + \angle DCE = 180^{\circ}$，$\therefore\angle DCE = 180^{\circ}-\angle ACB-\angle ACD =180^{\circ}-60^{\circ}-40^{\circ} = 80^{\circ}$。$\therefore\angle B = \angle DCE = 80^{\circ}$
