2. (2024春·儀征期末)在綜合與實(shí)踐課上,老師讓同學(xué)們以“三角板與平行線(xiàn)”為主題開(kāi)展數(shù)學(xué)活動(dòng)。已知直線(xiàn)$AB$,$CD$,$AB// CD$,在直角三角板$EFG$中,$\angle FEG= 90^{\circ}$,$\angle EGF= 60^{\circ}$。
(1)小明將三角板按如圖①方式擺放,點(diǎn)$G在CD$上,邊$GF與AB交于點(diǎn)H$,若$\angle FHA= 80^{\circ}$,則$\angle EGD$的度數(shù)是______;
(2)小亮將三角板按如圖②方式擺放,點(diǎn)$F$,$G分別在AB$,$CD$上,$\angle FEG的平分線(xiàn)與\angle FGC的平分線(xiàn)交于點(diǎn)M$,若$\angle EGD= 4\angle BFE$,求$\angle M$的度數(shù);
(3)小穎將圖②中的三角板進(jìn)行適當(dāng)轉(zhuǎn)動(dòng),點(diǎn)$F$,$G仍然分別在AB$,$CD$上,如圖③,再將$\angle DGE沿邊GE$翻折,邊$GD的對(duì)應(yīng)邊GN與AB交于點(diǎn)N$,小穎給出下列兩個(gè)結(jié)論:
①$\angle CGN+\angle BFE$的值不變;②$\frac{\angle CGN}{\angle BFE}$的值不變。
其中只有一個(gè)是正確的,你認(rèn)為哪個(gè)是正確的?請(qǐng)說(shuō)明理由。

答案:2. (1)$40^{\circ}$ 點(diǎn)撥:∵$\angle FHA = 80^{\circ}$,$AB // CD$,
∴$\angle CGH = \angle AHF = 80^{\circ}$.
∵$\angle FGE = 60^{\circ}$,∴$\angle EGD = 180^{\circ} - 60^{\circ} - 80^{\circ} = 40^{\circ}$.
(2)解:如答圖,過(guò)點(diǎn)$E$作$EK // AB$.
∵$AB // CD$,∴$AB // EK // CD$,
∴$\angle BFE = \angle KEF$,$\angle KEG = \angle EGD$.
∵$\angle FEK + \angle KEG = \angle FEG = 90^{\circ}$,
∴$\angle BFE + \angle EGD = 90^{\circ}$.
∵$\angle EGD = 4\angle BFE$,
∴$\angle BFE = 18^{\circ}$,$\angle EGD = 72^{\circ}$.
∵$\angle FGE = 60^{\circ}$,∴$\angle FGC = 180^{\circ} - 60^{\circ} - 72^{\circ} = 48^{\circ}$.
∵$EM$平分$\angle FEG$,$GM$平分$\angle FGC$,
∴$\angle FEM = \frac{1}{2} × 90^{\circ} = 45^{\circ}$,$\angle MGC = \frac{1}{2} × 48^{\circ} = 24^{\circ}$,
∴$\angle KEM = 45^{\circ} - 18^{\circ} = 27^{\circ}$,
同理可得:$\angle M = \angle KEM + \angle MGC = 27^{\circ} + 24^{\circ} = 51^{\circ}$.
(3)解:②$\frac{\angle CGN}{\angle BFE}$的值不變是正確的.理由如下:
設(shè)$\angle DGE = \angle NGE = x^{\circ}$,∴$\angle CGN = 180^{\circ} - 2x^{\circ}$.
同理可得:$\angle BFE + \angle DGE = \angle FEG = 90^{\circ}$,
∴$\angle BFE = 90^{\circ} - x^{\circ}$,
∴$\frac{\angle CGN}{\angle BFE} = \frac{180^{\circ} - 2x^{\circ}}{90^{\circ} - x^{\circ}} = 2$,
$\angle CGN + \angle BFE = 180^{\circ} - 2x^{\circ} + 90^{\circ} - x^{\circ} = 270^{\circ} - 3x^{\circ}$,
∴①$\angle CGN + \angle BFE$的值變化,②$\frac{\angle CGN}{\angle BFE}$的值不變.