22. (6分)已知$AM// CN$，$B$为平面内一点，$AB⊥ BC$于点$B$.
(1)如图1，直接写出$∠ A$和$∠ C$之间的数量关系:$\underline{\hspace{5em}}$;
(2)如图2，过点$B$作$BD⊥ AM$于点$D$，求证:$∠ ABD = ∠ C$;
(3)如图3，在(2)的条件下，点$E,F$在$DM$上，连接$BE,BF,CF$，$BF$平分$∠ DBC$，$BE$平分$∠ ABD$，若$∠ FCB + ∠ NCF = 180°$，$∠ BFC = 3∠ DBE$，求$∠ EBC$的度数.

22. 【点拨】本题考查平行线的判定与性质,余角和补角的性质,角平分线的定义.
【解析】(1)$\because AM // CN$,$\therefore ∠ C = ∠ AOB$.
$\because AB ⊥ BC$,$\therefore ∠ A + ∠ AOB = 90°$,$\therefore ∠ A + ∠ C = 90°$.
故答案为$∠ A + ∠ C = 90°$.
(2)证明:如题图2,过点B向右作$BG // DM$.
$\because BD ⊥ AM$,$\therefore BD ⊥ BG$,即$∠ ABD + ∠ ABG = 90°$.
又$\because AB ⊥ BC$,$\therefore ∠ CBG + ∠ ABG = 90°$,$\therefore ∠ ABD = ∠ CBG$.
$\because AM // CN$,$BG // AM$,$\therefore CN // BG$,
$\therefore ∠ C = ∠ CBG$,$\therefore ∠ ABD = ∠ C$.
(3)如题图3,过点B向右作$BG // DM$.
$\because BF$平分$∠ DBC$,$BE$平分$∠ ABD$,
$\therefore ∠ DBF = ∠ CBF$,$∠ DBE = ∠ ABE$.
由(2)得$∠ ABD = ∠ CBG$,$\therefore ∠ ABF = ∠ GBF$.设$∠ DBE = α$,$∠ ABF = β$,则$∠ ABE = α$,$∠ ABD = 2α = ∠ CBG$,$∠ GBF = β = ∠ AFB$,$∠ BFC = 3∠ DBE = 3α$,$\therefore ∠ AFC = 3α + β$.
$\because ∠ AFC + ∠ NCF = 180°$,$∠ FCB + ∠ NCF = 180°$,
$\therefore ∠ FCB = ∠ AFC = 3α + β$.
在$△ BCF$中,由$∠ CBF + ∠ BFC + ∠ BCF = 180°$得$(2α + β) + 3α + (3α + β) = 180°$①.
由$AB ⊥ BC$可得$β + β + 2α = 90°$②.
联立①②解得$α = 15°$,$\therefore ∠ ABE = 15°$,
$\therefore ∠ EBC = ∠ ABE + ∠ ABC = 15° + 90° = 105°$.

23. (6分)已知,直线$AB // CD$,点$E,F$分别在直线$AB,CD$上,$H$是直线$AB$与$CD$外一点,连接$HE,HF$.
(1)如图1,若$∠ CFH=120°,∠ H=120°$,求$∠ BEH$的度数;
(2)如图2,$∠ BEH$的平分线的反向延长线交$∠ CFH$的平分线于点$N$,猜想$∠ N$与$∠ H$的数量关系,并说明理由;
(3)如图3,若$∠ EHF=120°,∠ BEH=n∠ PEH,∠ CFH=n∠ HFQ$,点$P,H,Q$在同一直线上,直接写出$∠ Q - ∠ P$的值.(用含$n$的式子表示)

23. 【点拨】本题考查平行线的判定和性质,角平分线的定义.
【解析】(1)如图,过点H作$HG // AB$.
$\because AB // CD$,$\therefore GH // CD$,
$\therefore ∠ 1 + ∠ CFH = 180°$.
$\because ∠ CFH = 120°$,$\therefore ∠ 1 = 60°$.
$\because ∠ FHE = 120°$,$\therefore ∠ 2 = 60°$.
$\because HG // AB$,$\therefore ∠ BEH = ∠ 2 = 60°$.
(2)$∠ H = 180° - 2∠ N$.理由如下:
如图,过点N作$NQ // AB$,过点H作$HP // DC$.
$\because EM$平分$∠ BEH$,$FN$平分$∠ CFH$,
$\therefore$ 设$∠ 3 = ∠ 4 = α$,$∠ 6 = ∠ 7 = β$.
$\because AB // CD$,$\therefore AB // CD // NQ // PH$,
$\therefore ∠ 5 = ∠ 3 = α$,$∠ 8 = ∠ BEH = 2α$,$∠ QNF = ∠ 6 = β$,$∠ PHF = 180° - ∠ CFH = 180° - 2β$,
$\therefore ∠ ENF = ∠ QNF - ∠ 5 = β - α$,$∠ EHF = ∠ 8 + ∠ PHF = 2α + 180° - 2β = 180° - 2(β - α)$,
$\therefore ∠ EHF = 180° - 2∠ ENF$,即$∠ H = 180° - 2∠ N$.
(3)如图,过点P作$PK // AB$,过点H作$HL // AB$,过点Q作$QR // AB$.
$\because AB // CD$,$\therefore AB // CD // PK // HL // QR$.
设$∠ PEH = α$,$∠ HFQ = β$,则$∠ BEH = nα$,$∠ CFH = nβ$.
$\because PK // QR$,$\therefore ∠ KPQ = ∠ RQP$.
$\because AB // PK$,$\therefore ∠ BEP = ∠ KPE = (n-1)α$.
$\because QR // CD$,$\therefore ∠ CFQ = ∠ RQF = (n-1)β$.
$\because ∠ PQF = ∠ RQP + ∠ RQF$,$∠ EPQ = ∠ EPK + ∠ QPK$,
$\therefore ∠ PQF - ∠ EPQ = ∠ RQF - ∠ EPK = (n-1)(β - α)$.
$\because AB // HL // CD$,$\therefore ∠ EHL = ∠ BEH$,$∠ LHF = ∠ HFD$.
$\because ∠ EHF = ∠ EHL + ∠ LHF = 120°$,$\therefore ∠ BEH + ∠ HFD = 120°$,即$nα + 180° - nβ = 120°$,
$\therefore β - α = \dfrac{60°}{n}$,$\therefore ∠ PQF - EPQ = \dfrac{(n-1)60°}{n}$,
即$∠ Q - ∠ P = \dfrac{(n-1)60°}{n}$.