《2025年零障碍导教导学案九年级数学全一册人教版》

15. 如图,AB为⊙O的直径,点C,D在⊙O上,AC与OD交于点E,AE = EC,OE = ED,连接BC,CD. 求证：
(1)$\triangle AOE\cong\triangle CDE$；
证明:在$\triangle AOE$和$\triangle CDE$中，
$\begin{cases} AE = CE, \\ \angle AEO = \angle CED, \\ OE = DE, \end{cases}$
$\therefore \triangle AOE \cong \triangle CDE$(SAS).
(2)四边形OBCD是菱形.
证明:$\because \triangle AOE \cong \triangle CDE$,
$\therefore OA = CD,\angle AOE = \angle D$.
$\therefore$OB//CD.
$\because OA = OB$,
$\therefore$OB=CD.
$\therefore$四边形OBCD为平行四边形.
$\because$OB = OD.
$\therefore$四边形OBCD是菱形.

答案： 证明:
(1)在$\triangle AOE$和$\triangle CDE$中，
$\begin{cases} AE = CE, \\ \angle AEO = \angle CED, \\ OE = DE, \end{cases}$
$\therefore \triangle AOE \cong \triangle CDE(SAS)$.
(2)$\because \triangle AOE \cong \triangle CDE$,
$\therefore OA = CD,\angle AOE = \angle D$.
$\therefore OB // CD$.
$\because OA = OB$,
$\therefore OB = CD$.
$\therefore$四边形OBCD为平行四边形.
$\because OB = OD$.
$\therefore$四边形OBCD是菱形.

16. 如图,以△ABC的BC边上一点O为圆心的圆经过A,B两点,且与BC边交于点E,D为BE的下半圆弧的中点,连接AD交BC于点F,AC = FC.
(1)求证：AC是⊙O的切线；
(2)已知⊙O的半径R = 4,EF = 3,求DF的长.

答案：

(1)证明:如图,连接OA,OD,

$\because D$为$BE$的下半圆弧的中点，
$\therefore OD \perp BE$.
$\therefore \angle D + \angle DFO = 90^{\circ}$.
$\because AC = FC$,
$\therefore \angle CAF = \angle CFA$.
$\because \angle CFA = \angle DFO$,
$\therefore \angle CAF = \angle DFO$.
$\because OA = OD$,
$\therefore \angle OAD = \angle D$.
$\therefore \angle OAD + \angle CAF = 90^{\circ}$,
即$\angle OAC = 90^{\circ}$.
$\therefore OA \perp AC$.
$\because OA$是$\odot O$的半径，
$\therefore AC$是$\odot O$的切线.
(2)解:$\because \odot O$的半径$R = 4$，$EF = 3$，$\therefore OF = 1$.
在$Rt\triangle ODF$中，$OD = 4$，
$OF = 1$，
$\therefore DF = \sqrt{4^{2} + 1^{2}} = \sqrt{17}$.