8. 一题多解 [2026合肥四十二中段考] 如图,在$Rt△ ABC$中,$∠ C = 90°$,点$D$在$AC$上,$∠ DBC = ∠ A$.若$AC=4,\cos A=\dfrac{4}{5}$,则$BD$的长度为 ()


A.$\dfrac{9}{4}$
B.$\dfrac{12}{5}$
C.$\dfrac{15}{4}$
D.$4$

通解 $\because ∠ C=90°,AC=4,\cos A=\frac{4}{5},\therefore AB = \frac{AC}{\cos A}=5$,$\therefore BC = \sqrt{AB^2 - AC^2} = 3$. $\because ∠ DBC = ∠ A$,$\therefore \cos ∠ DBC = \frac{BC}{BD} = \cos A = \frac{4}{5}$,$\therefore BD = 3 × \frac{5}{4} = \frac{15}{4}$.
另解 $\because ∠ C=90°,AC=4,\cos A=\frac{4}{5}=\frac{AC}{AB},\therefore AB=5$,$\therefore BC = \sqrt{AB^2 - AC^2} = 3$. $\because ∠ DBC = ∠ A$,$∠ DCB = ∠ BCA$,$\therefore △ ABC ∽ △ BDC$,$\therefore \frac{AC}{BC} = \frac{AB}{BD}$,即$\frac{4}{3} = \frac{5}{BD}$,$\therefore BD = \frac{15}{4}$.

9. 教材P140例2变式[2024宣城宣州区期末]如图,若△ABC和△DEF的面积分别为S₁,S₂,则S₁:S₂= ()


A.5:8
B.8:5
C.1:1
D.2:7

解题思路:



过点$A$作$AM ⊥ BC$于点$M$,得$S_1 = \frac{1}{2}BC · AM = \frac{1}{2}BC · AB · \sin 40° = 20\sin 40°$
过点$D$作$DN ⊥ FE$交$FE$的延长线于点$N$,得$S_2 = \frac{1}{2}EF · DN = \frac{1}{2}EF · DE · \sin(180° - 140°) = 20\sin 40°$
$\therefore S_1:S_2 = 1:1$

10. [2026深圳实验中学期中]如图，$△ ABC$的顶点B，C的坐标分别是$(1,0),(0,\sqrt{3})$，且$∠ ABC = 90°, ∠ A = 30°$,则顶点A的坐标是。

过点$A$作$AG ⊥ x$轴于点$G$. $\because$ 点$B,C$的坐标分别是$(1,0),(0,\sqrt{3})$,$\therefore OC=\sqrt{3},OB=1$,$\therefore BC = \sqrt{1^2 + (\sqrt{3})^2} = 2$.
$\because ∠ ABC=90°,∠ BAC=30°$,$\therefore AB = \frac{BC}{\tan 30°} = \frac{2}{\frac{\sqrt{3}}{3}} = 2\sqrt{3}$.
$\because ∠ ABG + ∠ CBO = 90°$,$∠ BCO + ∠ CBO = 90°$,$\therefore ∠ ABG = ∠ BCO$. 又$\because ∠ AGB = ∠ BOC = 90°$,$\therefore △ AGB ∽ △ BOC$,
$\therefore \frac{AG}{OB} = \frac{BG}{OC} = \frac{AB}{BC}$,即$\frac{AG}{1} = \frac{BG}{\sqrt{3}} = \frac{2\sqrt{3}}{2}$,$\therefore AG=\sqrt{3},BG=3$,
$\therefore OG = OB + BG = 1 + 3 = 4$,$\therefore$ 顶点$A$的坐标是$(4,\sqrt{3})$.

11. [2026马鞍山段考] 如图，点E在矩形ABCD的边CD上，将△ADE沿AE折叠，点D恰好落在边BC上的点F处，若BC=10，sin∠AFB=$\frac{4}{5}$，则DE=。

解题思路:由矩形的性质得$∠ B = ∠ C = ∠ D = 90°$,$AD=BC=10$,$CD=AB$. 由折叠的性质得$AF=AD=10$,$FE=DE$. 由$\sin ∠ AFB = \frac{4}{5}$得$AB=8$,则$CD=AB=8$. 在$\mathrm{Rt}△ABF$中,由勾股定理,得$BF=6$,则$CF=BC-BF=4$. 在$\mathrm{Rt}△EFC$中,由勾股定理,易得$4^2 + (8-DE)^2 = DE^2$,解方程即可.
$\because$ 四边形$ABCD$是矩形,$BC=10$,$\therefore ∠ B = ∠ C = ∠ D = 90°$,$AD=BC=10$,$CD=AB$. 由折叠的性质,得$AF=AD=10$,$FE=DE$. 在$\mathrm{Rt}△ABF$中,$\sin ∠ AFB = \frac{AB}{AF} = \frac{4}{5}$,$\therefore AB = \frac{4}{5}AF = \frac{4}{5} × 10 = 8$,$\therefore CD=AB=8$. 在$\mathrm{Rt}△ABF$中,由勾股定理,得$BF = \sqrt{AF^2 - AB^2} = \sqrt{10^2 - 8^2} = 6$,$\therefore CF = BC - BF = 10 - 4 = 4$. 在$\mathrm{Rt}△EFC$中,由勾股定理,得$EF^2 = CF^2 + EC^2$,$\because FE=DE$,$CE=8-DE$,$\therefore 4^2 + (8-DE)^2 = DE^2$,解得$DE=5$.

12. ▶一题多解 [2025上海徐汇区期中]如图，已知在△ABD中，AC⊥BD，BC=8，CD=4，$\cos∠ ABC=\frac{4}{5}$，BF为AD边上的中线。
(1)求AC的长；
(2)求$\tan∠ FBD$的值。

解:(1) $\because AC ⊥ BD$,$\cos ∠ ABC = \frac{BC}{AB} = \frac{4}{5}$,$BC=8$,$\therefore AB=10$.
在$\mathrm{Rt}△ACB$中,由勾股定理,得$AC = \sqrt{AB^2 - BC^2} = \sqrt{10^2 - 8^2} = 6$,即$AC$的长为6.
(2)通解 如图,连接$CF$,过点$F$作$FE ⊥ BD$,垂足为$E$.
$\because BF$为$AD$边上的中线,即$F$为$AD$的中点,
$\therefore CF = \frac{1}{2}AD = FD$.
在$\mathrm{Rt}△ACD$中,由勾股定理,得$AD = \sqrt{AC^2 + CD^2} = \sqrt{6^2 + 4^2} = 2\sqrt{13}$,$\therefore CF = \sqrt{13}$.
$\because △ CFD$为等腰三角形,$FE ⊥ CD$,
$\therefore CE = \frac{1}{2}CD = 2$.
在$\mathrm{Rt}△EFC$中,$EF = \sqrt{CF^2 - CE^2} = 3$,
$\therefore \tan ∠ FBD = \frac{FE}{BE} = \frac{FE}{BC + CE} = \frac{3}{10}$.
另解 过点$F$作$FE ⊥ BD$,垂足为$E$.
$\because BF$为$AD$边上的中线,
$\therefore F$是$AD$的中点.
$\because FE ⊥ BD$,$AC ⊥ BD$,
$\therefore FE // AC$,$\therefore FE$是$△ ACD$的中位线,
$\therefore FE = \frac{1}{2}AC = 3$,$CE = \frac{1}{2}CD = 2$,
在$\mathrm{Rt}△BFE$中,$\tan ∠ FBD = \frac{FE}{BE} = \frac{FE}{BC + CE} = \frac{3}{10}$.

13. 运算能力 阅读下面材料：
小红遇到这样一个问题：如图1，在四边形ABCD中，∠A = ∠C = 90°，∠D = 60°，AB = 4√3，BC = √3，求AD的长。
小红发现，延长AB与DC，相交于点E，通过构造Rt△ADE能够使问题得到解决（如图2）。
(1)请回答：AD的长为；
(2)参考小红思考问题的方法，解决问题：
如图3，在四边形ABCD中，tan A = $\frac{1}{2}$，∠B = ∠C = 135°，AB = 9，CD = 3，求BC和AD的长。

解:(1)6
如题图2,在$△ ADE$中,$\because ∠ A=90°,∠ D=60°$,$\therefore ∠ E=30°$.
在$\mathrm{Rt}△BEC$中,$∠ BCE=90°$,$∠ E=30°$,$BC=\sqrt{3}$,$\therefore BE=2BC=2\sqrt{3}$,$\therefore AE=AB+BE=4\sqrt{3}+2\sqrt{3}=6\sqrt{3}$. 在$\mathrm{Rt}△ADE$中,$∠ A=90°$,$∠ E=30°$,$AE=6\sqrt{3}$,$\therefore AD=AE · \tan E = 6\sqrt{3} × \frac{\sqrt{3}}{3} = 6$.
(2)如图,延长$AB$与$DC$,相交于点$E$.
$\because ∠ ABC = ∠ BCD = 135°$,
$\therefore ∠ EBC = ∠ ECB = 45°$,
$\therefore BE=CE$,$∠ E=90°$.
设$BE=CE=x(x＞0)$,则$BC=\sqrt{2}x$,$AE=9+x$,$DE=3+x$.
在$\mathrm{Rt}△ADE$中,$∠ E=90°$,$\tan A = \frac{1}{2}$,
$\therefore \frac{DE}{AE} = \frac{1}{2}$,即$\frac{3+x}{9+x} = \frac{1}{2}$,$\therefore x=3$,
经检验,$x=3$是方程的解.
$\therefore BC=3\sqrt{2}$,$AE=12$,$DE=6$,
$\therefore AD = \sqrt{AE^2 + DE^2} = \sqrt{12^2 + 6^2} = 6\sqrt{5}$.