《2025年启东中学作业本九年级数学上册苏科版连淮专版》

9.如图,⊙O的弦AB,CD的延长线相交于点P,且AB= CD,求证：PB= PD.

答案：
证明：连接 $AC$，如答图。

$\because AB = CD$，$\therefore \overset{\frown}{AB} = \overset{\frown}{CD}$，$\therefore \overset{\frown}{AB} + \overset{\frown}{BD} = \overset{\frown}{CD} + \overset{\frown}{BD}$，
即 $\overset{\frown}{AD} = \overset{\frown}{BC}$，$\therefore \angle A = \angle C$，$\therefore PA = PC$，
$\therefore PA - AB = PC - CD$，
即 $PB = PD$。

10.(2024·赣榆区月考)如图,AB为⊙O的直径,CD为弦,CD⊥AB于点E,连接DO并延长交⊙O于点F,连接AF交CD于点G,连接AC,且AC//DF.
(1)求证：CG= AG；
(2)若AB= 12,求∠CAO的度数和DG的长.

答案：

(1) 证明：$\because AC// DF$，$\therefore \angle CDF = \angle ACD$。
$\because \angle CAF = \angle CDF$，
$\therefore \angle ACD = \angle CAF$，$\therefore AG = CG$。
(2) 解：如答图，连接 $CO$。

$\because AB\perp CD$，$\therefore \overset{\frown}{AC} = \overset{\frown}{AD}$，$CE = DE$。
$\because \angle DCA = \angle CAF$，$\therefore \overset{\frown}{AD} = \overset{\frown}{CF}$，
$\therefore \overset{\frown}{AC} = \overset{\frown}{AD} = \overset{\frown}{CF}$，$\therefore \angle AOD = \angle AOC = \angle COF$。
$\because DF$ 是直径，$\therefore \angle AOD = \angle AOC = \angle COF = 60^{\circ}$。
$\because OA = OC$，$\therefore \triangle AOC$ 是等边三角形，
$\therefore AC = AO = \frac{1}{2}AB = 6$，$\angle CAO = 60^{\circ}$。
$\because CE\perp AO$，$\therefore AE = EO = 3$，$\angle ACD = 30^{\circ}$，
$\therefore CE = 3\sqrt{3} = DE$。
$\because AG^{2} = GE^{2} + AE^{2}$，$\therefore AG^{2} = (3\sqrt{3} - AG)^{2} + 3^{2}$，
$\therefore AG = 2\sqrt{3}$，$\therefore GE = \sqrt{3}$，
$\therefore DG = DE + GE = 4\sqrt{3}$。

11.(2024·姑苏区期中)如图,在四边形ABCD中,∠B= ∠D,AB= CD,AB与DC不平行,过点A作AE//DC,交△ABC的外接圆⊙O于点E,连接CE,OA.求证：
(1)四边形ADCE为平行四边形；
(2)AO平分∠BAE.

答案：

(1) 证明：$\because \angle B = \angle E$，$\angle B = \angle D$，$\therefore \angle E = \angle D$。
$\because AE// CD$，$\therefore \angle E + \angle ECD = 180^{\circ}$，
$\therefore \angle D + \angle ECD = 180^{\circ}$，$\therefore CE// AD$，
$\therefore$ 四边形 $ADCE$ 为平行四边形。
(2) 过点 $O$ 作 $OG\perp AB$ 于点 $G$，$OH\perp AE$ 于点 $H$，如答图。
$\because$ 四边形 $ADCE$ 为平行四边形，$\therefore AE = CD$。
$\because AB = CD$，$\therefore AE = AB$。
$\because \angle AHO = \angle AGO = 90^{\circ}$，
$\therefore AH = \frac{1}{2}AE$，$AG = \frac{1}{2}AB$，$\therefore AH = AG$。
在 $Rt\triangle AHO$ 与 $Rt\triangle AGO$ 中，$\begin{cases}AO = AO, \\ AH = AG,\end{cases}$
$\therefore Rt\triangle AHO\cong Rt\triangle AGO(HL)$，
$\therefore \angle HAO = \angle GAO$，$\therefore AO$ 平分 $\angle BAE$。