《2025年名师帮同步学案九年级数学全一册人教版》

5. 如图,在半径为5的$\odot O$中,半径$OD\perp弦AB于点C$,连接$AO并延长交\odot O于点E$,连接$EC$,$EB$. 若$CD = 2$,求$EC$的长.

解:$\because \odot O$ 的半径为 5，$\therefore OA = OD = 5$，
$\because CD = 2$，$\therefore OC = OD - CD = 3$，
$\because OD \perp AB$，
$\therefore AC = BC = \sqrt{OA^2 - OC^2} = \sqrt{5^2 - 3^2} = 4$，
$\because OA = OE$，
$\therefore OC$ 是 $\triangle ABE$ 的中位线，
$\therefore BE = 2OC = 6$，
$\therefore EC = \sqrt{BC^2 + BE^2} = \sqrt{4^2 + 6^2} = 2\sqrt{13}$.

答案： 解:$\because \odot O$ 的半径为 5，$\therefore OA = OD = 5$，
$\because CD = 2$，$\therefore OC = OD - CD = 3$，
$\because OD \perp AB$，
$\therefore AC = BC = \sqrt{OA^2 - OC^2} = \sqrt{5^2 - 3^2} = 4$，
$\because OA = OE$，
$\therefore OC$ 是 $\triangle ABE$ 的中位线，
$\therefore BE = 2OC = 6$，
$\therefore EC = \sqrt{BC^2 + BE^2} = \sqrt{4^2 + 6^2} = 2\sqrt{13}$.

6. 如图,在$\odot O$中,$P是弦AB$的中点,$P'是弦A'B'$的中点.
(1) 若$AB = A'B'$,求证：$OP = OP'$；
(2) 若$OP = OP'$,求证：$AB = A'B'$.

答案：
证明:
(1)连接$OA$，$OA'$，则 $OA = OA'$.

$\because P$ 是弦 $AB$ 的中点，$P'$ 是弦 $A'B'$ 的中点，
$\therefore AP = \frac{1}{2}AB$，$A'P' = \frac{1}{2}A'B'$，$OP \perp AB$，$OP' \perp A'B'$.
$\because AB = A'B'$，$\therefore AP = A'P'$.
在 $Rt\triangle OAP$ 和 $Rt\triangle OA'P'$ 中，$\begin{cases} OA = OA' \\ AP = A'P' \end{cases}$，
$\therefore Rt\triangle OAP \cong Rt\triangle OA'P'$ (HL).
$\therefore OP = OP'$.
(2)由
(1) 知 $OP \perp AB$，$OP' \perp A'B'$，$OA = OA'$.
在 $Rt\triangle OAP$ 和 $Rt\triangle OA'P'$ 中，
$\begin{cases} OA = OA' \\ OP = OP' \end{cases}$，
$\therefore Rt\triangle OAP \cong Rt\triangle OA'P'$ (HL).
$\therefore AP = A'P'$.
又 $\because AB = 2AP$，$A'B' = 2A'P'$，
$\therefore AB = A'B'$.