《2025年大联考单元期末测试卷九年级数学全一册沪科版》

23. 如图 1,在△ABC 中,∠ACB = 90°,AC = BC,点 E 为△ABC 的中线 BD 上的一点,将线段 AE 以点 E 为旋转中心逆时针旋转 90°得到线段 EF,EF 恰好经过点 C.
(1) 若∠CAF = α,求∠CBE 的度数(用含 α 的代数式表示).
(2) 如图 2,过点 C 作 CH // AE 交 AF 于点 H,连接 BH 交 EF 于点 G.
① 求证:AF = BH;
② 若 CF = 2,求 EG 的长.

答案： 23.解:
(1)$\because$点$D$为$AC$的中点，$\angle AEC = 90^{\circ}$，
$\therefore AD = DE = DC$，$\therefore \angle DAE = \angle AED$.
$\because AE = EF$，$\therefore \angle EAF = \angle F = 45^{\circ}$，
$\therefore \angle EAD = 45^{\circ} - \alpha$，
$\therefore \angle DEA = \angle EAD = 45^{\circ} - \alpha$.
$\because \angle EDC = 90^{\circ} - 2\alpha$，$\angle BCA = 90^{\circ}$，
$\therefore \angle CBE = 2\alpha$.(4分)
(2)①证明:$\because CH // AE$，
$\therefore \angle FCH = \angle FEA = \angle BCA = 90^{\circ}$，
$\therefore \angle CHF = \angle EAF = \angle F = 45^{\circ}$，
$\therefore CH = CF$.(6分)
$\because \angle BCH = \angle ACH + \angle BCA$，
$\angle ACF = \angle ACH + \angle FCH$，
$\therefore \angle BCH = \angle ACF$.(7分)
在$\triangle ACF$和$\triangle BCH$中，$\because \begin{cases} BC = AC, \\\angle BCH = \angle ACF, \\CH = CF, \end{cases}$
$\therefore \triangle ACF \cong \triangle BCH(SAS)$，$\therefore AF = BH$.(9分)
②由$\triangle ACF \cong \triangle BCH$，得$\angle CAF = \angle CBH$.
又$\because \angle CBE = 2\angle CAF$，$\therefore \angle CBE = 2\angle CBH$，
$\therefore \angle CBH = \angle EBG$，$\therefore \angle CAF = \angle EBG$.
$\because DE = DC$，
$\therefore \angle DEC = \angle DCE$，即$\angle BEG = \angle ACF$，
$\therefore \triangle BEG \backsim \triangle ACF$.(12分)
由$BC = 2DC = 2DE$，可设$BC = 2x$，
则$AC = BC = 2x$，$CD = DE = x$，
$BD = \sqrt{BC^{2} + DC^{2}} = \sqrt{5}x$，$BE = (\sqrt{5} - 1)x$，
$\therefore \frac{BE}{BC} = \frac{\sqrt{5} - 1}{2}$，
$\therefore \frac{EG}{CF} = \frac{BE}{AC} = \frac{BE}{BC} = \frac{\sqrt{5} - 1}{2}$.
$\because CF = 2$，$\therefore EG = \sqrt{5} - 1$.(14分)