《2026年金考卷中考试题汇编45套数学山东专版》

20. (本小题满分9分)
如图,△ABC内接于⊙O,AB是⊙O的直径,点E在圆上,且BC = CE,过点C作CD⊥AE,交AE的延长线于D,DC与AB的延长线相交于点F.
(1) 求证:DF是⊙O的切线;
(2) 若BF = 2,tan∠FCB = $\frac{1}{2}$,求AD的长.

答案：

(1)证明:如图,连接$OC$.$\because BC = CE$,$\therefore \overset{\frown}{BC} = \overset{\frown}{CE}$,$\therefore \angle BAC = \angle CAD$(依据:等弧所对的圆周角相等).
$\because OA = OC$,$\therefore \angle OAC = \angle OCA$,$\therefore \angle CAD = \angle OCA$,$\therefore OC // AD$,$\therefore \angle OCF = \angle D = 90^{\circ}$.
$\because OC$是$\odot O$的半径,$\therefore DF$是$\odot O$的切线. (4分)

(2)如图,$\because AB$是$\odot O$的直径,$\therefore \angle ACB = 90^{\circ}$(依据:直径所对的圆周角是直角),$\therefore \angle OCB + \angle ACO = 90^{\circ}$.
由
(1)知$\angle OCF = 90^{\circ}$,$\therefore \angle BCF + \angle OCB = 90^{\circ}$,$\therefore \angle BCF = \angle ACO$.又$\because \angle OAC = \angle ACO$,$\therefore \angle BCF = \angle OAC$.
$\because \tan \angle BAC = \tan \angle FCB = \frac{1}{2}$,$\therefore \frac{BC}{AC} = \frac{1}{2}$.
$\because \angle F = \angle F$,$\angle BCF = \angle CAF$,$\therefore \triangle FBC \backsim \triangle FCA$,$\therefore \frac{FB}{FC} = \frac{FC}{FA} = \frac{BC}{AC} = \frac{1}{2}$,$\therefore FC = 4$,$FA = 8$,$\therefore AB = AF - BF = 8 - 2 = 6$,$\therefore OA = OC = 3$.
$\because OC // AD$,$\therefore \triangle FOC \backsim \triangle FAD$(点拨:“A”字型相似模型),$\therefore \frac{FO}{OC} = \frac{FA}{AD}$,$\therefore AD = \frac{FA · OC}{FO} = \frac{8 × 3}{2 + 3} = \frac{24}{5}$. (9分)

21. (本小题满分10分)
如图,直线l:y = x + b(b ≠ 0)与反比例函数y = $\frac{6}{x}$的图象交于A(m,3),D两点,与y轴、x轴分别交于点B,C.
(1) 求一次函数的表达式和点D的坐标.
(2) 请根据函数图象直接写出不等式$\frac{6}{x}$ ≥ x + b的解集.
(3) 请判断线段AB与CD的数量关系,并说明理由.

答案：

(1)把$A(m,3)$代入$y = \frac{6}{x}$,得$3 = \frac{6}{m}$,解得$m = 2$,$\therefore A(2,3)$,
将$A(2,3)$代入$y = x + b$,得$3 = 2 + b$,解得$b = 1$,$\therefore$一次函数的表达式为$y = x + 1$. (2分)
$\because$直线$y = x + 1$与反比例函数$y = \frac{6}{x}$的图象交于$A$,$D$两点,
$\therefore$联立$\begin{cases} y = x + 1, \\ y = \frac{6}{x} \end{cases}$解得$\begin{cases} x_{1} = -3, \\ y_{1} = -2, \end{cases} \begin{cases} x_{2} = 2, \\ y_{2} = 3, \end{cases}$$\therefore$点$D$的坐标为$(-3, -2)$. (4分)
(2)$x \leqslant -3$或$0 ＜ x \leqslant 2$. (6分)
(3)$AB = CD$.
理由:如图,过点$A$作$AM \perp y$轴于点$M$,过点$D$作$DN \perp x$轴于点$N$,
$\because$由
(1)知$A(2,3)$,$D(-3, -2)$,$\therefore AM = ND = 2$.
对于$y = x + 1$,当$y = 0$时,$x = -1$,当$x = 0$时,$y = 1$,$\therefore CO = BO = 1$.
又$\because \angle COB = 90^{\circ}$,$\therefore \angle OCB = \angle OBC = 45^{\circ}$,$\therefore \angle DCN = \angle ABM = 45^{\circ}$,$\therefore \triangle ABM \cong \triangle DCN$,$\therefore AB = CD$. (10分)