17. 已知,直线$AB// CD$,点$E$,$F分别在直线AB$,$CD$上,点$P是直线AB与CD$外一点,连接$PE$,$PF$。
(1)如图①,若$∠AEP= 45^{\circ }$,$∠DFP= 105^{\circ }$,求$∠EPF$的度数；
(2)如图②,过点$E作∠AEP的平分线EM交FP的延长线于点M$,$∠DFP的平分线FN交EM的反向延长线交于点N$,若$∠M与3∠N$互补,试探索直线$EP与直线FN$的位置关系,并说明理由；
(3)若点$P在直线AB的上方且不在直线EF$上,作$∠DFP的平分线FN交∠AEP的平分线EM所在直线于点N$,请直接写出$∠EPF与∠ENF$的数量关系。

解：(1)如图①，过点P作$PQ// AB$.
$\because AB// CD$，
$\therefore PQ// CD$.
$\therefore \angle QPE = \angle AEP = 45^{\circ}$，$\angle QPF = 180^{\circ} - \angle DFP = 180^{\circ} - 105^{\circ} = 75^{\circ}$.
$\therefore \angle EPF = \angle QPE + \angle QPF = 45^{\circ} + 75^{\circ} = 120^{\circ}$.
故$\angle EPF = 120^{\circ}$.

(2)$EP// FN$，
理由：如图②，$\because EM$平分$\angle AEP$，$FN$平分$\angle MFD$，
$\therefore \angle AEP = 2\angle 1$，$\angle MFD = 2\angle 3$.
$\because AB// CD$，
$\therefore \angle 3 = \angle 4$，
由(1)得，$\angle M = \angle 1 + \angle CFM = \angle 1 + (180^{\circ} - 2\angle 3) = \angle 1 + (180^{\circ} - 2\angle 4)$.
由三角形外角的性质可得，
$\angle N = \angle 4 - \angle 2 = \angle 4 - \angle 1$.
$\because \angle M$与$3\angle N$互补，
$\therefore \angle 1 + (180^{\circ} - 2\angle 4) + 3(\angle 4 - \angle 1) = 180^{\circ}$，
整理得，$\angle 4 = 2\angle 1 = \angle AEP$，
$\therefore EP// FN$.
(3)$\angle EPF + 2\angle ENF = 180^{\circ}$或$\angle EPF = 2\angle ENF - 180^{\circ}$.