《2025年高效精练九年级数学上册苏科版》

9. 如图,$\odot O$的半径$OD⊥$弦AB于点C,连接AO并延长交$\odot O$于点E,连接EC.若$AB=8,CD=2$,则EC的长为 (D)

A.$2\sqrt {15}$
B. 8
C.$2\sqrt {10}$
D.$2\sqrt {13}$

答案： D

10. 如图,已知在$△ABC$中,$AB=AC$.以AB为直径作$\odot O$,交BC于点D.若$∠BAC=40^{\circ }$,则$\widehat {AD}$的度数是140$^{\circ }$.

答案： 140

11. 如图,以AB为直径的$\odot O$经过$△ABC$的顶点C,AE,BE分别平分$∠BAC$和$∠ABC$,AE的延长线交$\odot O$于点D,连接BD.
(1)判断$△BDE$的形状,并证明你的结论;
(2)若$AB=10,BE=2\sqrt {10}$,求BC的长.

答案：

(1) $\triangle BDE$为等腰直角三角形. 证明: $\because AE$平分$\angle BAC$, $BE$平分$\angle ABC$, $\therefore \angle BAE = \angle CAD = \angle CBD$, $\angle ABE = \angle EBC$. $\because \angle BED = \angle BAE + \angle ABE$, $\angle DBE = \angle DBC + \angle CBE$, $\therefore \angle BED = \angle DBE$, $\therefore BD = ED$. $\because AB$为直径, $\therefore \angle ADB = 90^{\circ}$, $\therefore \triangle BDE$是等腰直角三角形.
(2) 如图, 连接$OC$, $CD$, $OD$, $OD$交$BC$于点$F$. $\because \angle DBC = \angle CAD = \angle BAD = \angle BCD$, $\therefore BD = DC$. $\because OB = OC$, $\therefore OD$垂直平分$BC$. $\because \triangle BDE$是等腰直角三角形, $BE = 2\sqrt{10}$, $\therefore BD = 2\sqrt{5}$. $\because AB = 10$, $\therefore OB = OD = 5$. 设$OF = t$, 则$DF = 5 - t$. 在$Rt\triangle BOF$和$Rt\triangle BDF$中, $5^{2} - t^{2} = (2\sqrt{5})^{2} - (5 - t)^{2}$, 解得$t = 3$, $\therefore BF = 4$, $\therefore BC = 8$.

12. 如图,AC是$\odot O$的弦,$AC=5$,点B是$\odot O$上的一个动点,且$∠ABC=45^{\circ }$,若点M,N分别是AC,BC的中点,则MN的最大值是______$\frac{5\sqrt{2}}{2}$.

答案： $\frac{5\sqrt{2}}{2}$

13. 如图,已知等腰直角三角形ABC,点P是斜边BC上一点(不与B,C重合),PE是$△ABP$的外接圆$\odot O$的直径.
(1)求证:$△APE$是等腰直角三角形;
(2)若$\odot O$的直径为2,求$PC^{2}+PB^{2}$的值.

答案：

(1) $\because AB = AC$, $\angle BAC = 90^{\circ}$, $\therefore \angle C = \angle ABC = 45^{\circ}$, $\therefore \angle AEP = \angle ABP = 45^{\circ}$. $\because PE$是直径, $\therefore \angle PAE = 90^{\circ}$, $\therefore \angle APE = \angle AEP = 45^{\circ}$. $\therefore AP = AE$, $\therefore \triangle PAE$是等腰直角三角形.
(2) 如图, 过点$P$作$PM \perp AC$于$M$, $PN \perp AB$于$N$, 则四边形$PMAN$是矩形, $\therefore PM = AN$. $\because \triangle PCM$, $\triangle PNB$都是等腰直角三角形, $\therefore PC = \sqrt{2}PM$, $PB = \sqrt{2}PN$, $\therefore PC^{2} + PB^{2} = 2(PM^{2} + PN^{2}) = 2(AN^{2} + PN^{2}) = 2PA^{2} = PE^{2} = 2^{2} = 4$. (也可以证明$\triangle ACP \cong \triangle ABE$, $\triangle PBE$是直角三角形)