13. 如图，已知$∠ 1=∠ BDC$，$∠ 2+∠ 3 = 180^{\circ}$.
(1)求证：$AD// CE$.
(2)若$DA$平分$∠ BDC$，$DA⊥ FE$于点$A$，$∠ FAB = 55^{\circ}$，求$∠ ABD$的度数.

13.（1）证明：$\because ∠1 = ∠BDC$，$\therefore AB// CD$，$\therefore ∠2 = ∠ADC$. $\because ∠2 + ∠3 = 180^{\circ }$，$\therefore ∠ADC + ∠3 = 180^{\circ }$，$\therefore AD// CE$. （2）$\because DA⊥FE$，$\therefore ∠DAF = 90^{\circ }$，$\therefore ∠2 = ∠DAF - ∠FAB = 90^{\circ } - 55^{\circ } = 35^{\circ }$. 由（1），知$AD// CE$，$\therefore ∠ADC = ∠2 = 35^{\circ }$. $\because DA$平分$∠BDC$，$∠1 = ∠BDC$，$\therefore ∠1 = ∠BDC = 2∠ADC = 70^{\circ }$，$\therefore ∠ABD = 180^{\circ } - ∠1 = 180^{\circ } - 70^{\circ } = 110^{\circ }$.

14. 如图，在$△ ABC$中，$AD⊥ BC$于点$D$，$BE$平分$∠ ABC$，已知$∠ EBC = 32^{\circ}$，$∠ AEB = 70^{\circ}$.
(1)求证：$∠ BAD:∠ CAD = 1:2$.
(2)若$F$为线段$BC$上的任意一点，当$△ EFC$为直角三角形时，求$∠ BEF$的度数.

14.（1）证明：$\because BE$平分$∠ABC$，$\therefore ∠ABC = 2∠EBC = 64^{\circ }$. $\because AD⊥BC$，$\therefore ∠ADB = ∠ADC = 90^{\circ }$，$\therefore ∠BAD = 90^{\circ } - 64^{\circ } = 26^{\circ }$. $\because ∠C + ∠EBC = ∠AEB$，$\therefore ∠C = ∠AEB - ∠EBC = 70^{\circ } - 32^{\circ } = 38^{\circ }$，$\therefore ∠CAD = 90^{\circ } - 38^{\circ } = 52^{\circ } = 2∠BAD$，$\therefore ∠BAD:∠CAD = 1:2$. （2）分两种情况：①如图1, $∠EFC = 90^{\circ }$，$\therefore ∠BEF = 90^{\circ } - ∠EBC = 90^{\circ } - 32^{\circ } = 58^{\circ }$；②如图2, $∠FEC = 90^{\circ }$，$\therefore ∠EFC = 90^{\circ } - 38^{\circ } = 52^{\circ }$，$\therefore ∠BEF = ∠EFC - ∠EBC = 52^{\circ } - 32^{\circ } = 20^{\circ }$. 综上所述，$∠BEF$的度数为58°或20°.
　　　　　　　　　　　　　　　　　　　　　　　　　　　　 　　

15. (1)如图1，$AB// CD$，则$∠ B+∠ D$$∠ E$.(填“$＞$”“$＜$”或“$=$”)
(2)写出(1)中命题的逆命题，判断逆命题的真假并说明理由.
(3)如图2，已知$AB// CD$，在$∠ ACD$的平分线上取两个点$M$、$N$，使得$∠ AMN=∠ ANM$，求证：$∠ CAM=∠ BAN$.

15.（1）= 解析：如图1，过点E作$EF// AB$，则$EF// AB// CD$，$\therefore ∠B = ∠BEF$，$∠D = ∠DEF$，$\therefore ∠B + ∠D = ∠BEF + ∠DEF = ∠BED$.
　　　　　　　　　　　　　　　　　　　　　　　　　　　　
（2）逆命题为：若$∠B + ∠D = ∠BED$，则$AB// CD$. 该逆命题为真命题. 理由如下：如图1，过点E作$EF// AB$，则$∠B = ∠BEF$. $\because ∠B + ∠D = ∠BED$，$∠BEF + ∠DEF = ∠BED$，$\therefore ∠D = ∠DEF$，$\therefore EF// CD$. $\because EF// AB$，$\therefore AB// CD$.
（3）证明：如图2，过点N作$NG// AB$，交AM于点G，则$NG// AB// CD$，$\therefore ∠BAN = ∠ANG$，$∠GNC = ∠NCD$. $\because ∠AMN$是$△ACM$的一个外角，$\therefore ∠AMN = ∠ACM + ∠CAM$. 又$\because ∠AMN = ∠ANM$，$∠ANM = ∠ANG + ∠GNC$，$\therefore ∠ACM + ∠CAM = ∠ANG + ∠GNC$，$\therefore ∠ACM + ∠CAM = ∠BAN + ∠NCD$. $\because CN$平分$∠ACD$，$\therefore ∠ACM = ∠NCD$，$\therefore ∠CAM = ∠BAN$.