《2025年实验班提优训练九年级数学上册人教版》

7. 如图,$△ABC内接于\odot O$,AB为$\odot O$的直径,延长AC到点G,使得$CG= CB$,连接GB.过点C作$CD// GB$,交AB于点F,交$\odot O$于点D,过点D作$DE// AB$,交GB的延长线于点E.
(1)求证:DE与$\odot O$相切;
(2)若$AC= 4,BC= 2$,求BE的长.

(1)连接OD,
∵AB为$\odot O$的直径,∴$\angle ACB=\angle BCG = 90^{\circ}$.
∵$CG = CB$,∴$\triangle BCG$为等腰直角三角形,
∴$\angle G=\angle CBG = 45^{\circ}$.
∵$CD// GB$,∴$\angle ACD=\angle G = 45^{\circ}$,$\angle BCD=\angle CBG = 45^{\circ}$,
∴$\angle AOD = 2\angle ACD = 90^{\circ}$.
∵$DE// AB$,∴$\angle ODE=\angle AOD = 90^{\circ}$,即$OD\perp DE$.
又点D在$\odot O$上,∴OD为$\odot O$的半径，
∴DE为$\odot O$的切线，即DE与$\odot O$相切.
(2)$\frac{5\sqrt{2}}{3}$

答案：
(1)连接OD,
∵AB为$\odot O$的直径,
∴$\angle ACB=\angle BCG = 90^{\circ}$.
∵$CG = CB$,
∴$\triangle BCG$为等腰直角三角形,
∴$\angle G=\angle CBG = 45^{\circ}$.
∵$CD// GB$,
∴$\angle ACD=\angle G = 45^{\circ}$,$\angle BCD=\angle CBG = 45^{\circ}$,
∴$\angle AOD = 2\angle ACD = 90^{\circ}$.
∵$DE// AB$,
∴$\angle ODE=\angle AOD = 90^{\circ}$,即$OD\perp DE$.
又点D在$\odot O$上,
∴OD为$\odot O$的半径，
∴DE为$\odot O$的切线，即DE与$\odot O$相切.
(2)由
(1)可知,$\angle ACB = 90^{\circ}$,$\angle ACD=\angle BCD = 45^{\circ}$,$\angle AOD = 90^{\circ}$.
在$Rt\triangle ABC$中,$AC = 4$,$BC = 2$,
由勾股定理,得$AB=\sqrt{AC^{2}+BC^{2}} = 2\sqrt{5}$.
∴$OA = OB = OD=\sqrt{5}$.
∵$CD// GB$,$AC = 4$,$BC = CG = 2$,
∴$AF:BF = AC:CG = 4:2 = 2:1$.
设$BF = k$,$AF = 2k$,
∴$AB = AF + BF = 3k = 2\sqrt{5}$,
∴$k=\frac{2\sqrt{5}}{3}$,
∴$AF = 2k=\frac{4\sqrt{5}}{3}$,
∴$OF = AF - OA=\frac{4\sqrt{5}}{3}-\sqrt{5}=\frac{\sqrt{5}}{3}$.
在$Rt\triangle ODF$中,$OD=\sqrt{5}$,$OF=\frac{\sqrt{5}}{3}$,
由勾股定理,得$DF=\sqrt{OD^{2}+OF^{2}}=\frac{5\sqrt{2}}{3}$.
∵$CD// GB$,$DE// AB$,
∴四边形DEBF为平行四边形,
∴$BE = DF=\frac{5\sqrt{2}}{3}$.

8. 如图,$△ABC内接于\odot O,∠B= 60^{\circ }$,CD是$\odot O$的直径,点P是CD延长线上的一点,且$AP= AC.$
(1)求证:PA是$\odot O$的切线;
(2)若$AB= 4+\sqrt {3},BC= 2\sqrt {3}$,求$\odot O$的半径.

答案：

(1)如图,连接OA.
∵$\angle B = 60^{\circ}$,
∴$\angle AOC = 2\angle B = 120^{\circ}$.
∵$OA = OC$,
∴$\angle OAC=\angle OCA = 30^{\circ}$.
∵$AP = AC$,
∴$\angle P=\angle ACP = 30^{\circ}$,
∴$\angle OAP=\angle AOC-\angle P = 90^{\circ}$,
∴$OA\perp PA$.（三角形外角定理）
又OA是$\odot O$的半径,
∴PA是$\odot O$的切线.
(2)如图,过点C作$CE\perp AB$于点E.
∵在$Rt\triangle BCE$中,$\angle B = 60^{\circ}$,$BC = 2\sqrt{3}$,
∴$\angle BCE = 30^{\circ}$,
∴$BE=\frac{1}{2}BC=\sqrt{3}$,$CE=\sqrt{BC^{2}-BE^{2}} = 3$.
（含$30^{\circ}$角的直角三角形中,较短直角边是斜边的一半）
∵$AB = 4+\sqrt{3}$,
∴$AE = AB - BE = 4$,
∴在$Rt\triangle ACE$中,$AC=\sqrt{AE^{2}+CE^{2}} = 5$,
∴$AP = AC = 5$.
∵在$Rt\triangle PAO$中,$\angle P = 30^{\circ}$,
∴$OP = 2OA$.
由勾股定理,得$OA^{2}+AP^{2}=OP^{2}$,
∴$OA^{2}+5^{2}=(2OA)^{2}$,解得$OA=\frac{5\sqrt{3}}{3}$(负值已舍去),
∴$\odot O$的半径为$\frac{5\sqrt{3}}{3}$.

9. 如图,O为$△ABC$的边AB上的一点,以点O为圆心,AO的长为半径作圆,交AB于点D,过点A作$AE// BC$,交$\odot O$于点E.
(1)如图(1),连接DE,若$∠B= 45^{\circ }$,则$∠ADE= $______.
(2)如图(2),连接CD,交$\odot O$于点F,连接AF,EF,且$∠EFA= ∠ACB.$
①求证:AC为$\odot O$的切线;
②若$AC= 6,BC= 10,AF= EF$,求$△BCD$的面积.

答案：

(1)$45^{\circ}$
(2)①
∵$AE// BC$,
∴$\angle EAB=\angle B$.
∵$\angle EAB=\angle EFD$,
∴$\angle B=\angle EFD$.
∵AD为$\odot O$的直径,
∴$\angle AFD = 90^{\circ}$,
∴$\angle EFD+\angle EFA = 90^{\circ}$.
∵$\angle EFA=\angle ACB$,
∴$\angle B+\angle ACB = 90^{\circ}$,
∴$\angle BAC = 90^{\circ}$,
∴$OA\perp AC$.
又OA为$\odot O$的半径,
∴AC为$\odot O$的切线.
②如图,连接ED,并延长交CB于点G.
∵$AF = FE$,
∴$\angle FEA=\angle FAE$.
∵四边形AFDE为圆内接四边形,
∴$\angle FAE+\angle FDE = 180^{\circ}$.
又$\angle FDE+\angle CDG = 180^{\circ}$,
∴$\angle CDG=\angle FAE$.
∵$\angle ADF=\angle AEF$,
∴$\angle CDG=\angle ADF$.
∵AD为$\odot O$的直径,
∴$\angle DEA = 90^{\circ}$.
∵$EA// CB$,
∴$\angle DGC = 90^{\circ}$,
∴$\angle DGC=\angle CAD$.
又$CD = CD$,$\angle CDG=\angle CDA$,
∴$\triangle CGD\cong\triangle CAD(AAS)$,
∴$DG = AD$,$AC = GC = 6$.
在$Rt\triangle ABC$中,
∵$\angle CAB = 90^{\circ}$,$BC = 10$,$AC = 6$,
∴$AB=\sqrt{BC^{2}-AC^{2}} = 8$.
设$AD = x$,则在$Rt\triangle BDG$中,$DG = AD = x$,$BD = BA - AD = 8 - x$,$BG = BC - GC = 4$.
由勾股定理,得$BG^{2}+DG^{2}=BD^{2}$,
∴$4^{2}+x^{2}=(8 - x)^{2}$,
解得$x = 3$,即$DG = 3$,
∴$S_{\triangle BCD}=\frac{1}{2}BC\cdot DG=\frac{1}{2}\times10\times3 = 15$.

10. (2024·东营中考)如图,$△ABC内接于\odot O$,AB是$\odot O$的直径,点E在$\odot O$上,点C是$\widehat {BE}$的中点,$AE⊥CD$,垂足为D,DC的延长线交AB的延长线于点F.
(1)求证:CD是$\odot O$的切线;
(2)若$CD= \sqrt {3},∠ABC= 60^{\circ }$,求线段AF的长.

(1)连接OC,
∵点C是$\overset{\frown}{BE}$的中点,∴$\overset{\frown}{BC}=\overset{\frown}{CE}$,∴$\angle BAC=\angle CAE$.
∵$OC = OA$,∴$\angle OCA=\angle OAC$,
∴$\angle OCA=\angle CAD$,∴$OC// AD$.
∵$AE\perp CD$,∴$OC\perp DF$.
∵OC是$\odot O$的半径,∴CD是$\odot O$的切线.
(2)∵AB是$\odot O$的直径,∴$\angle ACB = 90^{\circ}$.
∵$\angle ABC = 60^{\circ}$,∴$\angle BAC = 30^{\circ}$,
∴$\angle CAD=\angle BAC = 30^{\circ}$.
∵$\angle D = 90^{\circ}$,$CD=\sqrt{3}$,∴$AC = 2CD = 2\sqrt{3}$,
∴$AD=\sqrt{AC^{2}-CD^{2}}=\sqrt{(2\sqrt{3})^{2}-(\sqrt{3})^{2}} = 3$.
在$Rt\triangle ADF$中,∵$\angle F = 180^{\circ}-\angle D-\angle BAD = 30^{\circ}$,
∴$AF = 2AD =$6.

答案：
(1)连接OC,
∵点C是$\overset{\frown}{BE}$的中点,
∴$\overset{\frown}{BC}=\overset{\frown}{CE}$,
∴$\angle BAC=\angle CAE$.
∵$OC = OA$,
∴$\angle OCA=\angle OAC$,
∴$\angle OCA=\angle CAD$,
∴$OC// AD$.
∵$AE\perp CD$,
∴$OC\perp DF$.
∵OC是$\odot O$的半径,
∴CD是$\odot O$的切线.
(2)
∵AB是$\odot O$的直径,
∴$\angle ACB = 90^{\circ}$.
∵$\angle ABC = 60^{\circ}$,
∴$\angle BAC = 30^{\circ}$,
∴$\angle CAD=\angle BAC = 30^{\circ}$.
∵$\angle D = 90^{\circ}$,$CD=\sqrt{3}$,
∴$AC = 2CD = 2\sqrt{3}$,
∴$AD=\sqrt{AC^{2}-CD^{2}}=\sqrt{(2\sqrt{3})^{2}-(\sqrt{3})^{2}} = 3$.
在$Rt\triangle ADF$中,
∵$\angle F = 180^{\circ}-\angle D-\angle BAD = 30^{\circ}$,
∴$AF = 2AD = 6$.