19. (6分)如图,$\triangle ABC$的周长为27,$AC= 9$,$BC边上的中线AE= 6$,$\triangle ABE$的周长为19,求$AB$的长.
答案：解:设 $AB = x,BE = y$。$\because AE$ 是 $BC$ 边上的中线，$\therefore BC = 2BE = 2y$。由题意得 $x + 9 + 2y = 27$，$x + 6 + y = 19$，解得 $x = 8,y = 5$。$\therefore AB$ 的长为 8。
20. (6分)如图,在$\triangle ABC$中,$∠B= 25^{\circ }$,$∠BAC= 31^{\circ }$,过点$A作BC$边上的高,交$BC的延长线于点D$,$CE平分∠ACD$,交$AD于点E$.
求：(1)$∠ACD$的度数；
(2)$∠AEC$的度数.
答案：解:(1) $\because \angle ACD = \angle B + \angle BAC,\angle B = 25^{\circ},\angle BAC = 31^{\circ},\therefore \angle ACD = 25^{\circ} + 31^{\circ} = 56^{\circ}$。
(2) $\because AD\bot BD,\therefore \angle D = 90^{\circ}$。由(1)知 $\angle ACD = 56^{\circ},\because CE$ 平分 $\angle ACD$，$\therefore \angle ECD = \frac{1}{2}\angle ACD = 28^{\circ}$，$\therefore \angle AEC = \angle ECD + \angle D = 28^{\circ} + 90^{\circ} = 118^{\circ}$。
21. (6分)如图,$CE是\triangle ABC的外角∠ACD$的平分线,且$CE交BA的延长线于点E$.求证：$∠BAC= ∠B+2∠E$.
答案：证明: $\because CE$ 平分 $\angle ACD,\therefore \angle ACE = \angle DCE$。$\because \angle BAC = \angle ACE + \angle E,\therefore \angle BAC = \angle DCE + \angle E$。$\because \angle DCE = \angle B + \angle E,\therefore \angle BAC = \angle B + 2\angle E$。
22. (10分)如图,直线$a// b$,点$A$,$D在直线a$上,点$C$,$B在直线b$上,连接$AC$,$AB$,$CD$,$AB与CD交于点E$,其中$AB平分∠DAC$,$∠ACB= 80^{\circ }$,$∠BED= 110^{\circ }$.
求：(1)$∠ABC$的度数；(2)$∠ACD$的度数.
答案：解:(1) $\because a// b$，$\therefore \angle DAB = \angle ABC,\angle DAC + \angle ACB = 180^{\circ}$。$\because \angle ACB = 80^{\circ},\therefore \angle DAC = 100^{\circ}$。$\because AB$ 平分 $\angle DAC,\therefore \angle DAB = \angle CAB = 50^{\circ}$，$\therefore \angle ABC = \angle DAB = 50^{\circ}$。
(2) $\because \angle AEC = \angle BED = 110^{\circ},\angle EAC = 50^{\circ}$，$\therefore \angle ACD = 180^{\circ} - 110^{\circ} - 50^{\circ} = 20^{\circ}$。
