《2025年南方新课堂金牌学案九年级数学上册人教版》

例2 如图,$\triangle ABC$是直角三角形,$\angle ABC = 90^{\circ}$,以$AB为直径的\odot O交AC于点E$,点$D是BC$边的中点,连接$DE$.
(1)求证：$DE与\odot O$相切；

(2)若$\odot O的半径为\sqrt{3}$,$DE = 3$,求$AE$.

答案： 解
(1)证明：连接$OE$,$BE$.
$\because AB$是直径,$\therefore BE \perp AC$.
$\because D是BC$的中点,$\therefore DE = DB$.
$\therefore \angle DBE = \angle DEB$.
又$\because OE = OB$,$\therefore \angle OBE = \angle OEB$.
$\therefore \angle DBE + \angle OBE = \angle DEB + \angle OEB$,即$\angle ABD = \angle OED$.
$\because \angle ABC = 90^{\circ}$,$\therefore \angle OED = 90^{\circ}$.
$\therefore DE是\odot O$的切线.
(2)$\because AC = \sqrt{AB^{2} + BC^{2}} = \sqrt{(2\sqrt{3})^{2} + 6^{2}} = 4\sqrt{3}$,
$\therefore BE = \frac{AB \cdot BC}{AC} = \frac{2\sqrt{3} \cdot 6}{4\sqrt{3}} = 3$.
$\therefore AE = \sqrt{AB^{2} - BE^{2}} = \sqrt{12 - 9} = \sqrt{3}$.

1. 如图,$P为\odot O$外一点,$PA切\odot O于点A$,且$OP = 5$,$PA = 4$,则$OA$等于(B).

A.4
B.3
C.5
D.$\sqrt{41}$

答案： B

2. 如图,AB是$\odot O$的直径,点C在AB的延长线上,CD与$\odot O$相切于点D,连接AD,若$\angle A = 28^{\circ},$则$\angle C = $34°.

答案： 34°

3. 如图,已知$\odot O内切于\triangle ABC$,切点分别为$D$,$E$,$F$,若$\angle A = 50^{\circ}$,则$\angle EDF = $65°.

答案： 65°

4. 已知$\triangle ABC的周长为24$,面积为$48$,则它的内切圆的半径为4.

答案： 4