《2025年全品学练考九年级数学上册苏科版江苏专版》

8. 如图2-5-15,AB为$\odot O$的弦,$OD⊥AB$于点D,过点A作$\odot O$的切线交OD的延长线于点C.若$∠BOC= 57^{\circ }$,则$∠C$的度数为(D)

A.$43^{\circ }$
B.$37^{\circ }$
C.$35^{\circ }$
D.$33^{\circ }$

答案： D

9. 如图2-5-16,PA,PB是$\odot O$的切线,A,B为切点,点C在$\odot O$上,且$∠ACB= 55^{\circ }$,则$∠APB$的度数为70$^{\circ }$.

答案： 70

10. (2023南京期中)如图2-5-17,在四边形ABCD中,AC,BD相交于点E,且$AB= AC= AD$,经过A,C,D三点的$\odot O$交BD于点F,连接CF.
(1)求证:$CF= BF$;
(2)若$CD= CB$,求证:CB是$\odot O$的切线.

答案：
证明:
(1)
∵AB = AC,
∴∠ACB = ∠ABC.
∵AB = AD,
∴∠ADB = ∠ABD.
又
∵∠ADB = ∠ACF,
∴∠ACF = ∠ABD,
∴∠ACB - ∠ACF = ∠ABC - ∠ABD,即∠BCF = ∠CBF,
∴CF = BF.
(2)如图,连接CO并延长交⊙O于点G,再连接GF.

∵CG为⊙O的直径,
∴∠GFC = 90°,
∴∠G + ∠GCF = 90°.
∵∠CDB = ∠G,
∴∠CDB + ∠GCF = 90°.
∵CD = CB,
∴∠CDB = ∠CBD.
∵CF = BF,
∴∠BCF = ∠CBD,
∴∠BCF = ∠CDB,
∴∠BCF + ∠GCF = 90°,
∴∠BCG = 90°,
∴CG⊥BC.
又
∵OC是⊙O的半径，
∴CB是⊙O的切线

11. 核心素养推理能力如图2-5-18,$\odot O为\triangle ABC$的外接圆,$AC= AB$,作直线$AD// BC,CD⊥AD$于点D.
(1)求证:AD为$\odot O$的切线;
(2)如图②,CD交$\odot O$于点E,连接BE,过点A作$AG⊥BE$于点F,交BC于点G,若$AD= 3,CD= 4$,求FG的长.

答案：
解:
(1)证明:如图,连接OA,OB,OC.

在△OAC和△OAB中,
$\begin{cases}AC = AB,\\AO = AO,\\CO = BO,\end{cases}$
∴△OAC≌△OAB.
∴∠OAC = ∠OAB,即AO平分∠BAC.
∴AO⊥BC.
又
∵AD//BC,
∴AD⊥AO.
∵AO为⊙O的半径，
∴AD为⊙O的切线
(2)
∵$\overset{\frown}{AE}=\overset{\frown}{AE}$,
∴∠ACD = ∠ABF.
在△ADC和△AFB中,
$\begin{cases}\angle ADC = \angle AFB = 90^{\circ},\\\angle ACD = \angle ABF,\\AC = AB,\end{cases}$
∴△ADC≌△AFB(AAS).
∴AF = AD = 3,BF = CD = 4,∠BAG = ∠CAD.
∵AD//BC,
∴∠CAD = ∠ACB.
∵AC = AB,
∴∠ACB = ∠ABC.
∴∠BAG = ∠ABC.
∴AG = BG.
设FG = x,
在Rt△BFG中,FG = x,BF = 4,BG = AG = x + 3,
∴FG² + BF² = BG²,
即x² + 4² = (x + 3)²,
解得x = $\frac{7}{6}$.
∴FG = $\frac{7}{6}$.