16. (8分)(2026·广安中考)如图,△ABC内接于⊙O,AD是⊙O的直径,过点A作射线AE交BC的延长线于点E,使∠EAB=∠ECA.
(1)求证:AE是⊙O的切线;
(2)过点C作CF//AE交AB于点F,若AF=2,AB=9,求AC的长.

(1)证明:如图,连接$DC$,$\because ∠ ECA=∠ B+∠ BAC$,$∠ EAB=∠ CAE+∠ BAC$,$∠ ECA=∠ EAB$,$\therefore ∠ B+∠ BAC=∠ CAE+∠ BAC$,$\therefore ∠ B=∠ CAE. \because AD$是$\odot O$的直径,$\therefore ∠ ACD=90°$,$\therefore ∠ D+∠ DAC=90°. \because ∠ B=∠ D$,$\therefore ∠ CAE=∠ D$,$\therefore ∠ CAE+∠ DAC=90°$,$\therefore DA⊥ AE$,$\therefore AE$是$\odot O$的切线. (2)解:$\because CF// AE$,$\therefore ∠ FCA=∠ CAE$,$∠ CAE=∠ B$,$\therefore ∠ FCA=∠ B.$ 在$△ ACF$和$△ ABC$中,$\because ∠ FCA=∠ B$,$∠ CAF=∠ BAC$,$\therefore △ ACF∽ △ ABC$,$\therefore \dfrac{AC}{AB}=\dfrac{AF}{AC}$,$\therefore AC^2=AB· AF. \because AB=9$,$AF=2$,$\therefore AC^2=9× 2=18$,$\therefore AC=3\sqrt{2}.$

17. (8分)如图①所示的是古代一种可以远程攻击的投石车,图②是投石车投石过程中某时刻的示意图,GP是杠杆,弹袋挂在点G处,重锤挂在点P处,点A为支点,点D是水平底板BC上的一点,AD=AC=3米,CD=3.6米.

(1)投石车准备时,点G恰好与点B重合,此时AG和AC垂直,求AG的长;
(2)投石车投石瞬间,AP的延长线交线段DC于点E,若DE:CE=5:1,则此时点G到水平底板BC的距离为多少米?

17. (1)解:如图①,连接$AB$,过$A$点作$AF⊥ BC$于$F$,$\because AD=AC=3$米,$CD=3.6$米,$\therefore CF=DF=1.8$米,$\therefore AF=\sqrt{AC^2-CF^2}=2.4$米.$\because ∠ B+∠ ACB=90°$,$∠ CAF+∠ ACB=90°$,$\therefore ∠ B=∠ CAF. \because ∠ AFB=∠ AFC=90°$,$\therefore △ AFB∽ △ CFA$,$\therefore \dfrac{AF}{CF}=\dfrac{BF}{AF}$,$\therefore BF=2.4^2÷ 1.8=3.2$(米),$\therefore AB=\sqrt{AF^2+BF^2}=4$米,$\therefore AG$的长为4米. (2)解:过点$G$作$GF⊥ DC$于点$F$,过点$A$作$AH⊥ CD$于点$H$,则$∠ AHE=∠ GFE=90°$,如图②,$\because CD=3.6$米,$DE:CE=5:1$,$\therefore CE=0.6$米,$\therefore EH=1.8-0.6=1.2$(米),$\therefore AE=\sqrt{1.2^2+2.4^2}=\dfrac{6\sqrt{5}}{5}$(米).$\because ∠ AEH=∠ GEF$,$∠ EHA=∠ EFG=90°$,$\therefore △ EAH∽ △ EGF$,$\therefore \dfrac{AH}{GF}=\dfrac{AE}{EG}$,$\therefore \dfrac{2.4}{GF}=\dfrac{\dfrac{6\sqrt{5}}{5}}{\dfrac{6\sqrt{5}}{5}+4}$,解得$GF=\dfrac{8\sqrt{5}+12}{5}$,$\therefore$ 此时点$G$到水平底板$BC$的距离为$\dfrac{8\sqrt{5}+12}{5}$米.

18. (9分)(2025·徐州期末)在$△ ABC$中,点D在边AB上,若$CD^2=AD·DB$,则称点D是点C的“关联点”.
(1)如图①,在$△ ABC$中,若$∠ ACB=90°$,$CD⊥ AB$于点D,试说明:点D是点C的“关联点”;
(2)如图②,已知点D在线段AB上,用无刻度的直尺和圆规作一个$△ ABC$,使点D为点C的“关联点”.

19. (12分)(2026·眉山中考)如图,一次函数

18. (1)证明:$\because CD⊥ AB$,$\therefore ∠ CDA=∠ CDB=90°$,$\therefore ∠ A+∠ ACD=90°. \because ∠ ACB=90°$,$\therefore ∠ BCD+∠ ACD=90°$,$\therefore ∠ A=∠ BCD. \because ∠ CDA=∠ CDB=90°$,$\therefore △ ACD∽ △ CBD$,$\therefore \dfrac{CD}{BD}=\dfrac{AD}{CD}$,$\therefore CD^2=AD· DB$,$\therefore$ 点$D$是点$C$的“关联点”. (2)解:如图,$△ ABC$即为所求. 作法如下:作线段$AB$的垂直平分线,交$AB$于点$O$;以$O$为圆心,$OA$为半径作圆;过$D$作$DC⊥ AB$交$\odot O$于点$C$,连接$AC,BC$,可得$△ ABC$是直角三角形,结合(1)中的结论可得点$D$为点$C$的“关联点”.