24. (9分)已知$AB// CD$，点$M$在直线$AB$上，点$N,Q$在直线$CD$上，点$P$在直线$AB,CD$之间，$∠ AMP = ∠ PQN = α$，$PQ$平分$∠ MPN$.
(1)如图1，求$∠ MPQ$的度数；(用含$α$的式子表示)
(2)如图2，过点$Q$作$QE// PN$交$PM$的延长线于点$E$，过点$E$作$EF$平分$∠ PEQ$交$PQ$于点$F$.请你判断$EF$与$PQ$的位置关系，并说明理由；
(3)如图3，在(2)的条件下，连接$EN$，若$NE$平分$∠ PNQ$，请你判断$∠ NEF$与$∠ AMP$的数量关系，并说明理由.

24. 【点拨】本题考查平行线的性质和角平分线的定义.
【解析】(1)如题图1,过点P向右作$PF // AB$,则$PF // AB // CD$,
$\therefore ∠ FPM = ∠ PMA = α$,$∠ FPQ = ∠ PQN = α$,
$\therefore ∠ MPQ = ∠ MPF + ∠ FPQ = 2α$.
(2)$EF ⊥ PQ$.理由如下:
$\because PN // QE$,$\therefore ∠ NPE + ∠ QEP = 180°$.
又$\because PQ$平分$∠ MPN$,$EF$平分$∠ PEQ$,
$\therefore ∠ FPE + ∠ FEP = \dfrac{1}{2}∠ NPE + \dfrac{1}{2}∠ QEP = 90°$,
$\therefore ∠ EFP = 180° - ∠ FPE - ∠ FEP = 90°$,
$\therefore PF ⊥ EF$,$\therefore EF ⊥ PQ$.
(3)$∠ AMP = 2∠ NEF$.理由如下:
$\because ∠ NPQ = 2α$,$∠ PQN = α$,$\therefore ∠ PND = 180° - 3α$.
又$\because NE$平分$∠ PNQ$,$\therefore ∠ DNE = ∠ PNE = 90° - \dfrac{3}{2}α$.
由(2)知,$∠ PEF = 90° - 2α$.
又$\because PN // QE$,$\therefore ∠ NEQ = ∠ PNE = 90° - \dfrac{3}{2}α$,$∠ PEQ = 180° - 2∠ MPQ = 180° - 4α$,
$\therefore ∠ NEF = ∠ PEF - ∠ PEN = ∠ PEF - (∠ PEQ - ∠ NEQ) = ∠ PEF + ∠ NEQ - ∠ PEQ = 90° - 2α + (90° - \dfrac{3}{2}α) - (180° - 4α) = \dfrac{1}{2}α$,
$\therefore ∠ AMP = 2∠ NEF$.