《2025年天府前沿九年级数学上册北师大版》

9. (1)(2025·编写)如图,$\angle B= 90^{\circ}$,$\angle ACB= 30^{\circ}$,$AC= 2$,$AD= 4$,$DC= 2\sqrt{3}$,求证：$\triangle ABC \backsim \triangle ACD$.

(2)(2025·编写)如图,在$\triangle ABC和\triangle ADE$中,$\angle BAD= \angle CAE$,$\angle ABD= \angle ACE$. 求证：
①$AB \cdot AE= AC \cdot AD$；
②$\triangle ADE \backsim \triangle ABC$.

答案：
(1)[证明]$\because ∠B = 90^{\circ}, ∠ACB = 30^{\circ}, AC = 2,$
$\therefore AB = \frac{1}{2}AC = 1, BC = \frac{\sqrt{3}}{2}AC = \sqrt{3}. \because AD = 4, DC = 2\sqrt{3},$
$\therefore \frac{AB}{AC} = \frac{1}{2}, \frac{BC}{CD} = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}, \frac{AC}{AD} = \frac{2}{4} = \frac{1}{2},$
$\therefore \frac{AB}{AC} = \frac{BC}{CD} = \frac{AC}{AD}, \therefore △ABC \sim △ACD.$
(2)[证明]①$\because ∠BAD = ∠CAE, ∠ABD = ∠ACE,$
$\therefore △ABD \sim △ACE, \therefore \frac{AB}{AC} = \frac{AD}{AE}, \therefore AB \cdot AE = AC \cdot AD.$
②$\because ∠BAD = ∠CAE, \therefore ∠BAD + ∠DAC = ∠CAE + ∠DAC,$
即$∠BAC = ∠DAE$. 由①得$\frac{AB}{AC} = \frac{AD}{AE}, \therefore \frac{AB}{AD} = \frac{AC}{AE}, \therefore △ADE \sim △ABC.$

10. (2025·编写)如图,已知$AB // A^{\prime}B^{\prime}$,$\frac{B^{\prime}C^{\prime}}{BC}= \frac{OB^{\prime}}{OB}$,$\frac{A^{\prime}C^{\prime}}{AC}= \frac{OA^{\prime}}{OA}$. 求证：$\triangle ABC \backsim \triangle A^{\prime}B^{\prime}C^{\prime}$.

答案： [证明]$\because AB // A'B',$
$\therefore △OA'B' \sim △OAB,$
$\therefore \frac{A'B'}{AB} = \frac{OB'}{OB} = \frac{OA'}{OA}.$
$\because \frac{B'C'}{BC} = \frac{OB'}{OB}, \frac{A'C'}{AC} = \frac{OA'}{OA},$
$\therefore \frac{A'B'}{AB} = \frac{B'C'}{BC} = \frac{A'C'}{AC},$
$\therefore △ABC \sim △A'B'C'.$

11. (2025·编写)如图,在正方形网格中有5个三角形(三角形的顶点均在格点上)：①$\triangle ABC$,②$\triangle ADE$,③$\triangle AEF$,④$\triangle AFH$,⑤$\triangle AHG$,在②至⑤中,与①相似的三角形是______.(填序号)

答案： ②④

12. (2025·编写)如图,在$\triangle ABC$中,$\angle BAC= 45^{\circ}$,$\angle ACB= 30^{\circ}$,将$\triangle ABC绕点A顺时针旋转得到\triangle AB_{1}C_{1}$,当$C$,$B_{1}$,$C_{1}$三点共线时,旋转角为$\alpha$,连接$BB_{1}交AC于点D$,有以下结论：①$\triangle AC_{1}C$为等腰三角形；②$CA= CB_{1}$；③$\alpha=135^{\circ}$；④$\triangle AB_{1}D \backsim \triangle ACB_{1}$；⑤$\frac{AB}{B_{1}C}= \frac{\sqrt{6}-\sqrt{2}}{2}$. 其中正确结论的序号为______.

答案： ①②④⑤

13. (2025·编写)如图,在边长为4的正方形$ABCD$中,$E$,$F分别是BC$,$CD$的中点,$DE与AF交于点G$,$AF的中点为H$,连接$BG$,$DH$. 给出下列结论：①$AF \perp DE$；②$DG= \frac{8}{5}$；③$HD // BG$；④$\triangle ABG \backsim \triangle DHF$. 其中正确的结论有______.(请填上所有正确结论的序号)

答案： ①④

14. (2025·编写)如图,在正方形$ABCD$中,点$E$,$F分别在边AD$,$AB$上,且$AE= AF$,又$AG \perp BE于点G$. 求证：
(1)$\frac{AF}{AG}= \frac{BC}{BG}$；
(2)$GF \perp GC$.

答案：
(1)$\because AG$是$Rt△ABE$的斜边$BE$上的高,
$\therefore Rt△AGE \sim Rt△BGA, \therefore \frac{AE}{AG} = \frac{AB}{BG}.$
$\because AE = AF, AB = BC, \therefore \frac{AF}{AG} = \frac{BC}{BG}.$
(2)由$Rt△AGE \sim Rt△BGA$, 得$∠GAF = ∠GEA.$
$\because AD // BC, \therefore ∠GEA = ∠CBG, \therefore ∠GAF = ∠CBG;$
$\because \frac{AF}{AG} = \frac{BC}{BG}, \therefore △AGF \sim △BGC, \therefore ∠AGF = ∠BGC;$
$\because ∠AGF + ∠FGB = 90^{\circ}, \therefore ∠BGC + ∠FGB = 90^{\circ},$
$\therefore GF \perp GC.$