《2025年名师帮同步学案九年级数学全一册人教版》

5. 如图,以$□ ABCD的顶点A$为圆心,$AB为半径作⊙A$,$⊙A交AD$,$BC于点E$,$F$,延长$BA交⊙A于点G$,求证：$\overgroup{GE}= \overgroup{EF}$.

答案：
证明：如图，连接$AF$.

$\because AB = AF$，
$\therefore \angle B = \angle AFB$.
$\because$四边形$ABCD$为平行四边形，
$\therefore AD // BC$，
$\therefore \angle GAE = \angle B$，$\angle AFB = \angle EAF$，
$\therefore \angle GAE = \angle EAF$，
$\therefore \overarc{GE} = \overarc{EF}$.

6. (2024·潮南月考)如图,$OA$,$OB是⊙O$的半径,且$\overgroup{AC}= \overgroup{BC}$,弦$CM$,$CN分别经过OA$,$OB的中点D$,$E$. 求证：$CM = CN$.

答案：
证明：如图，过点$C$作直径$CF$.

$\because \overarc{AC} = \overarc{BC}$，$\therefore \angle AOC = \angle BOC$.
$\because$点$D$，$E$分别是$OA$，$OB$的中点，$OA = OB$，
$\therefore OD = OE$.
在$\triangle OCD$和$\triangle OCE$中，$\left\{ \begin{array} { l } { O D = O E }, \\ { \angle C O D = \angle C O E }, \\ { O C = O C }, \end{array} \right.$
$\therefore \triangle OCD \cong \triangle OCE ( S A S )$，
$\therefore \angle MCF = \angle NCF$，
$\therefore \overarc{FM} = \overarc{FN}$.
$\because CF$是$\odot O$的直径，
$\therefore \overarc{FMC} = \overarc{FNC}$，
$\therefore \overarc{MC} = \overarc{NC}$，
$\therefore CM = CN$.