《2025年大联考单元期末测试卷九年级数学全一册沪科版》

17. 如图,△ABC 的内心为 I,连接 AI 并延长交△ABC 的外接圆于点 D.
(1)求证:DI = DB.
(2)求证:点 D 是△BCI 的外心.

答案：
17.证明:
(1)如图,连接 BI.

$\because \triangle ABC$的内心为 I,
$\therefore \angle 1 = \angle 2$,$\angle 5 = \angle 6$.
$\because \angle 3 = \angle 1$,$\therefore \angle 3 = \angle 2$.
$\because \angle 4 = \angle 2 + \angle 6$,$\angle DBI = \angle 3 + \angle 5$,
$\therefore \angle 4 = \angle DBI$,$\therefore DI = DB$.(5 分)
(2)如图,连接 CD,IC.
$\because \angle 1 = \angle 2$,
$\therefore BD = CD$,
$\therefore$点 D 在 BC 的垂直平分线上.
$\because DI = DB$,$\therefore$点 D 在 IB 的垂直平分线上,
$\therefore$点 D 是$\triangle BCI$的外心.(10 分)

18. 如图,AB 为⊙O 的直径,点 C 为 BA 延长线上一点,CD 是⊙O 的切线,点 D 为切点,OF⊥AD 交 AD 于点 E,交 CD 于点 F.
(1)求证:∠AOF = ∠ADC.
(2)若 sin C = $\frac{1}{3}$,BD = 8,求 EF 的长.

答案：
18.解:
(1)证明:如图,连接 OD.

$\because AB$为$\odot O$的直径,
$\therefore \angle ADB = 90^{\circ}$,$\therefore AD\perp BD$.
$\because OF\perp AD$,$\therefore OF// BD$,$\therefore \angle AOF=\angle B$.(2 分)
$\because CD$是$\odot O$的切线,点 D 为切点,
$\therefore \angle CDO = 90^{\circ}$,
$\therefore \angle CDA+\angle ADO=\angle ADO+\angle BDO = 90^{\circ}$,
$\therefore \angle CDA=\angle BDO$.
$\because OD = OB$,$\therefore \angle ODB=\angle B$,
$\therefore \angle AOF=\angle ADC$.(5 分)
(2)$\because OF// BD$,$AO = OB$,$\therefore AE = DE$,
$\therefore OE=\frac{1}{2}BD=\frac{1}{2}×8 = 4$.
$\because \sin C=\frac{OD}{OC}=\frac{1}{3}$,$\therefore$设$OD = x$,$OC = 3x$,
$\therefore OB = x$,$\therefore CB = 4x$.(7 分)
$\because OF// BD$,$\therefore \triangle COF\backsim\triangle CBD$,$\therefore \frac{OC}{BC}=\frac{OF}{BD}$,
$\therefore \frac{3x}{4x}=\frac{OF}{8}$,$\therefore OF = 6$,$\therefore EF = OF - OE = 6 - 4 = 2$.(10 分)