《2025年学霸题中题九年级数学上册人教版》

12. 如图,在$\odot O$中,AB为直径,弦CD交AB于点P,且$OP = PC$,试猜想$\overset{\frown}{AD}与\overset{\frown}{CB}$之间的关系,并证明你的猜想.

猜想：$\overset{\frown}{AD}$与$\overset{\frown}{CB}$之间的关系是$\overset{\frown}{AD}=3\overset{\frown}{CB}$。
证明：连接OC和OD。$\because OC = OD$，$\therefore \angle D = \angle C$。$\because OP = PC$，$\therefore \angle C = \angle COP$，$\therefore \angle D = \angle C = \angle COP$。又$\because \angle AOD = \angle DPO + \angle D$，$\angle DPO = \angle C + \angle COP$，$\therefore \angle AOD = \angle C + \angle COP + \angle D = 3\angle COP$，$\therefore \overset{\frown}{AD}=3\overset{\frown}{CB}$。

答案： $\overset{\frown}{AD}=3\overset{\frown}{CB}$，证明：连接OC和OD。$\because OC = OD$，$\therefore \angle D = \angle C$。$\because OP = PC$，$\therefore \angle C = \angle COP$，$\therefore \angle D = \angle C = \angle COP$。又$\because \angle AOD = \angle DPO + \angle D$，$\angle DPO = \angle C + \angle COP$，$\therefore \angle AOD = \angle C + \angle COP + \angle D = 3\angle COP$，$\therefore \overset{\frown}{AD}=3\overset{\frown}{CB}$。

13. 如图①,在同圆或等圆中,如果两个圆心角对应的两条弧、两条弦或两条弦的弦心距中有一组量相等,那么它们对应的其余各组量也相等. 请直接运用圆心角、弧、弦、弦心距之间的关系解答下列问题.
如图②,O是$∠EPF$的平分线上一点,以点O为圆心的圆与角的两边分别交于点A,B,C,D.
(1)求证:$AB = CD$.
(2)若角的顶点P在圆上或圆内,上述结论还成立吗? 若不成立,请说明理由;若成立,请加以证明.

答案：

(1)证法一：如图①，过O作$OM\perp AB$于M，$ON\perp CD$于N，则$\angle OMB = \angle OND = 90^{\circ}$。$\because PO$平分$\angle EPF$，$\therefore OM = ON$，$\therefore AB = CD$。
证法二：如图①，过O作$OM\perp AB$于M，$ON\perp CD$于N，求出$ON = OM$，证明$\triangle OMB\cong\triangle OND(HL)$，推出$BM = DN$，根据垂径定理得出$AB = 2BM$，$CD = 2DN$，$\therefore AB = CD$。
(2)结论还成立，证明：如图②，当P在$\odot O$上时，$\because$由
(1)知，$BM = DN$，$AB = 2BM$，$CD = 2DN$，$\therefore AB = CD$；当P在$\odot O$内时，如图③，$\because$由
(1)知，$BM = DN$，$AB = 2BM$，$CD = 2DN$，$\therefore AB = CD$。

14. 如图,AB是$\odot O$的直径,点D,C在$\odot O$上,$∠DOC = 90^{\circ}$,$AC = 2$,$BD = 2\sqrt{2}$,则$\odot O$的半径为 ( )

A. $\sqrt{3}$
B. $\sqrt{5}$
C. $\sqrt{2}+1$
D. $\sqrt{10}$

答案：
D 解析：作半径$OE\perp AB$，交$\odot O$于点E，连接DE，作$BF\perp ED$，交ED的延长线于点F，连接BE，如图，$\because \angle DOC = 90^{\circ}$，$\angle BOE = \angle AOE = \angle BFE = 90^{\circ}$，$\therefore \angle DOE = \angle AOC$，$\therefore DE = AC = 2$。$\because \angle ODE = \frac{1}{2}\times(180^{\circ}-\angle DOE)$，$\angle ODB = \frac{1}{2}\times(180^{\circ}-\angle BOD)$，$\angle DOE + \angle BOD = \angle BOE = 90^{\circ}$，$\therefore \angle BDE = \angle ODE + \angle ODB = 135^{\circ}$，$\therefore \angle BDF = 45^{\circ}$。$\because BD = 2\sqrt{2}$，$\therefore DF = BF = 2$，$\therefore EF = 4$，$\therefore BE = \sqrt{2^{2}+4^{2}} = 2\sqrt{5}$。$\because \triangle BOE$为等腰直角三角形，$\therefore OB = \sqrt{10}$。

15. 如图,将半径为2,圆心角为$120^{\circ}$的扇形OAB绕点A逆时针旋转一定角度,使点O的对应点D落在$\overset{\frown}{AB}$上,点B的对应点为点C,连接BC,则BC的长度是____.

答案：
$2\sqrt{3}$ 解析：如图，连接OD，BD。由题意，得$OA = OD = AD$，$\therefore \triangle AOD$是等边三角形，$\therefore \angle ADO = \angle AOD = 60^{\circ}$。$\because \angle ADC = \angle AOB = 120^{\circ}$，$\therefore \angle ADO + \angle ADC = 180^{\circ}$，$\therefore O$，$D$，$C$三点共线。$\because \angle AOD = \angle DOB = 60^{\circ}$，$OD = OB$，$\therefore \triangle OBD$是等边三角形，$\therefore \angle BDO = 60^{\circ}$。$\because DC = DB$，$\therefore \angle DCB = \angle DBC = 30^{\circ}$，$\therefore \angle OBC = 90^{\circ}$，$\therefore BC = 2\sqrt{3}$。

16. 如图,CD是$\odot O$的直径,点A是半圆上的三等分点,点B是$\overset{\frown}{AD}$的中点,点P为直线CD上的一个动点,当$CD = 4$时,求:
(1)$AP + BP$的最小值;
(2)$AP - BP$的最大值.

答案：

(1)如图①，作点A关于CD的对称点$A'$，连接$A'B$，交CD于点P，此时$PA + PB$最小，连接OA，AP，OB，$OA'$，$AA'$。$\because$点A与$A'$关于CD对称，点A是半圆上的一个三等分点，$\therefore \angle A'OD = \angle AOD = 60^{\circ}$，$PA = PA'$。$\because$点B是$\overset{\frown}{AD}$的中点，$\therefore \angle BOD = 30^{\circ}$，$\therefore \angle A'OB = \angle A'OD + \angle BOD = 90^{\circ}$。又$OA = OA' = OB = 2$，$\therefore A'B = 2\sqrt{2}$，$\therefore AP + BP$的最小值即为$A'P + BP = A'B = 2\sqrt{2}$。


(2)由三角形的三边关系可知，当A，B，P三点共线时，$AP - BP$有最大值AB，如图②，连接AO，BO，AB，过点A作$AN\perp OB$，垂足为N。$\because CD$是$\odot O$的直径，点A是半圆上的三等分点，点B是$\overset{\frown}{AD}$的中点，$CD = 4$，$\therefore \angle AOB = 30^{\circ}$，$AN = \frac{1}{2}AO = 1$，$\therefore ON = \sqrt{3}$，$BN = 2 - \sqrt{3}$，$\therefore AP - BP$的最大值为$AB = \sqrt{1^{2}+(2 - \sqrt{3})^{2}} = \sqrt{8 - 4\sqrt{3}} = \sqrt{6 - 4\sqrt{3} + 2} = \sqrt{6} - \sqrt{2}$。