《2025年大联考单元期末测试卷九年级数学全一册沪科版》

20. 如图,△ABC 内接于⊙O,AO 的延长线交⊙O 于点 D,交 BC 于点 E,过点 D 作 DF // BC 交⊙O 于点 F,连接 CF,CD.
(1) 若 CF // AD,求证:AC = CE.
(2) 求证:点 O 到 AB 的距离等于 $\frac{1}{2}CF$ 的长.

答案：
20.证明:
(1)$\because CF // AD$，$\therefore \angle AEC = \angle ECF$；
$\because DF // BC$，$\therefore \angle ECF + \angle CFD = 180^{\circ}$，
$\therefore \angle AEC + \angle CFD = 180^{\circ}$.
$\because$四边形$ADFC$是$\odot O$的内接四边形，
$\therefore \angle EAC + \angle CFD = 180^{\circ}$，$\therefore \angle AEC = \angle EAC$，$\therefore AC = CE$.(5分)
(2)如图，过点$O$作$OG \perp AB$于点$G$，连接$BD$.

$\because DF // BC$，$\therefore \overset{\frown} {BD} = \overset{\frown} {CF}$，$\therefore BD = CF$.
$\because AD$是$\odot O$的直径，$\therefore \angle ABD = 90^{\circ}$.
$\because OG \perp AB$，$\therefore \angle AGO = 90^{\circ}$，
$\therefore \angle ABD = \angle AGO$，$\therefore OG // BD$.
$\because AO = OD$，$\therefore OG$是$\triangle ABD$的中位线，
$\therefore OG = \frac{1}{2}BD$，$\therefore OG = \frac{1}{2}CF$，
$\therefore$点$O$到$AB$的距离等于$\frac{1}{2}CF$的长.(10分)