《2025年全品学练考九年级数学上册人教版》

9. (2023德阳)将一副三角尺DOE与AOC叠放在一起,如图23-Y-8①,$∠O=90^{\circ },∠A=30^{\circ },∠E=45^{\circ },OD＞OC$.将三角尺DOE绕点O顺时针旋转$\alpha (0^{\circ }＜\alpha ＜90^{\circ })$到$\triangle D_{1}OE_{1}$的位置,使$OD_{1}// AC$,如图②.
(1)求α的值;
(2)如图③,继续将三角尺DOE绕点O顺时针旋转,使点E落在AC边上的点$E_{2}$处,点D落在点$D_{2}$处,设$E_{2}D_{2}$交$OD_{1}$于点G,$OE_{1}$交AC于点H,若G是$E_{2}D_{2}$的中点,试判断四边形$OHE_{2}G$的形状,并说明理由.

答案：
(1)$\alpha = 30^{\circ}$
(2)四边形$OHE_2G$是正方形.理由如下:
$\because \angle E_2OD_2 = 90^{\circ}, OD_2 = OE_2, G$是$E_2D_2$的中点,
$\therefore E_2G = OG, E_2G \perp OG$,
$\therefore \angle OGE_2 = 90^{\circ}$.
$\because OD_1 // AC$,
$\therefore \angle CE_2G = 180^{\circ} - \angle OGE_2 = 90^{\circ}$.
又$\because \angle HOG = 90^{\circ}$,
$\therefore$四边形$OHE_2G$是矩形.
又$\because E_2G = OG$,
$\therefore$四边形$OHE_2G$是正方形.

10. (2024北京)已知$∠MAN=\alpha (0^{\circ }＜\alpha ＜45^{\circ })$,点B,C分别在射线AN,AM上,将线段BC绕点B顺时针旋转$180^{\circ }-2\alpha $得到线段BD,过点D作AN的垂线交射线AM于点E.
(1)如图23-Y-9①,当点D在射线AN上时,求证:C是AE的中点;
(2)如图②,当点D在$∠MAN$内部时,作$DF// AN$,交射线AM于点F,用等式表示线段EF与AC的数量关系,并证明.

答案：
解:
(1)证明:如图①，连接$CD$.

由题意,得$BC = BD, \angle CBD = 180^{\circ} - 2\alpha$,
$\therefore \angle BDC = \angle BCD = \frac{1}{2}(180^{\circ} - \angle CBD) = \alpha$,
$\therefore \angle BDC = \angle A$,
$\therefore CA = CD$.
$\because DE \perp AN$,
$\therefore \angle 1 + \angle A = \angle 2 + \angle BDC = 90^{\circ}$,
$\therefore \angle 1 = \angle 2$,
$\therefore CD = CE$,
$\therefore CA = CE$,
$\therefore C$是$AE$的中点.
(2)$EF = 2AC$.证明如下:
如图②，在射线$AM$上取一点$H$，使得$BH = BA$，取$EF$的中点$G$，连接$DH, DG$.

$\because BH = BA$,
$\therefore \angle BAH = \angle BHA = \alpha$,
$\therefore \angle ABH = 180^{\circ} - 2\alpha = \angle CBD$,
$\therefore \angle ABC = \angle HBD$.
又$\because BC = BD$,
$\therefore \triangle ABC \cong \triangle HBD(SAS)$,
$\therefore AC = HD, \angle BHD = \angle A = \alpha$,
$\therefore \angle FHD = \angle BHA + \angle BHD = 2\alpha$.
$\because DF // AN$,
$\therefore \angle EFD = \angle A = \alpha, \angle EDF = \angle 3 = 90^{\circ}$.
$\because G$是$EF$的中点,
$\therefore GF = GD, EF = 2GD$,
$\therefore \angle GFD = \angle GDF = \alpha$,
$\therefore \angle HGD = 2\alpha$,
$\therefore \angle HGD = \angle FHD$,
$\therefore GD = HD$.
$\because AC = HD$,
$\therefore GD = AC$,
$\therefore EF = 2AC$.