10.特色 多解法「2024海南中考,★☆」如圖,AD是半圓O的直徑,點B,C在半圓上,且$\overset{\frown}{AB}= \overset{\frown}{BC}= \overset{\frown}{CD}$,點P在$\overset{\frown}{CD}$上,若∠PCB= 130°,則∠PBA等于( )
A.105°
B.100°
C.90°
D.70°
答案：B 如圖，連接OB，OC. ∵AD是半圓O的直徑，∴∠AOD = 180°. ∵$\overset{\frown}{AB}$ = $\overset{\frown}{BC}$ = $\overset{\frown}{CD}$，∴∠AOB = ∠BOC = ∠COD = 60°. ∵OA = OB = OC，∴△AOB和△BOC均是等邊三角形，∴∠ABO = ∠CBO = ∠BCO = 60°，∴∠ABC = ∠ABO + ∠CBO = 120°.
[解法一] ∵∠BPC是$\overset{\frown}{BC}$所對的圓周角，∠BOC是$\overset{\frown}{BC}$所對的圓心角，
∴∠BPC = $\frac{1}{2}$∠BOC = 30°.
∵∠PCB = 130°，∴∠PBC = 180° - ∠BPC - ∠PCB = 180° - 30° - 130° = 20°，
∴∠ABP = ∠ABC - ∠PBC = 120° - 20° = 100°.故選B.
[解法二] 如圖，連接OP. ∵OC = OP，∴△COP是等腰三角形，∵∠PCB = 130°，∴∠OPC = ∠OCP = ∠PCB - ∠BCO = 130° - 60° = 70°，∴∠COP = 180° - ∠OPC - ∠OCP = 180° - 70° - 70° = 40°，∴∠PBC = $\frac{1}{2}$∠COP = $\frac{1}{2}$×40° = 20°，∴∠PBA = ∠ABC - ∠PBC = 120° - 20° = 100°.故選B.
11.「2025浙江杭州蕭山期中,★☆」如圖,AB為⊙O的直徑,點C為圓上一點,若將劣弧AC沿弦AC翻折交AB于點D,連接CD,∠DCA= 44°,則∠BAC的度數(shù)為( )
A.23°
B.24°
C.25°
D.26°
答案：A 如圖，連接BC，∵AB是⊙O的直徑，∴∠ACB = 90°，∵∠DCA = 44°，
∴∠BCD = 90° - 44° = 46°，根據(jù)翻折的性質(zhì)，知$\overset{\frown}{ABC}$所對的圓周角為∠ADC，∵$\overset{\frown}{AC}$所對的圓周角為∠B，∴∠ADC + ∠B = 180°，∵∠ADC + ∠BDC = 180°，∴∠B = ∠BDC = $\frac{1}{2}$×(180° - 46°) = 67°，∴∠BAC = 90° - 67° = 23°.故選A.
12.「2024江蘇連云港中考,★☆」如圖,AB是圓的直徑,∠1、∠2、∠3、∠4的頂點均在AB上方的圓弧上,∠1、∠4的一邊分別經(jīng)過點A、B,則∠1+∠2+∠3+∠4=
90
°.
答案：答案 90
解析 ∵AB是圓的直徑，∴AB所對的弧是半圓，所對圓心角的度數(shù)為180°，∵∠1、∠2、∠3、∠4所對的弧的和為半圓，∴∠1 + ∠2 + ∠3 + ∠4 = $\frac{1}{2}$×180° = 90°.
13.「2024山西朔州懷仁期中,★☆」如圖,AC,BD是⊙O的兩條相交弦,∠ACB= ∠CDB= 60°,AC= 2$\sqrt{3}$,則⊙O的直徑是______.
答案：答案 4
解析 如圖，作直徑BM，連接CM，∴∠BCM = 90°，易知∠A = ∠D = ∠M = 60°，∵∠ACB = 60°，
∴∠ABC = 60°，∴△ABC是等邊三角形，∴BC = AC = 2$\sqrt{3}$. 在Rt△BCM中，∠M = 60°，∴∠CBM = 30°，∴BM = 2CM，∵BC2 + CM2 = BM2，∴(2$\sqrt{3}$)2 + CM2 = 4CM2，∴CM = 2（舍負），∴BM = 4，∴⊙O的直徑是4.
14.「2024浙江湖州南潯期末,★☆」如圖,△ABC中,AB= AC,以AB為直徑作⊙O,交BC邊于點D,交CA的延長線于點E,連接AD,DE.
(1)求證：BD= CD.
(2)若AB= 5,DE= 4,求AD的長.
(1)證明：∵AB是⊙O的直徑，∴∠ADB = 90°，∴AD⊥BD，又∵AB = AC，∴BD = CD.
(2)∵AB = 5，∴AC = AB = 5，∴∠B = ∠C. ∵∠B = ∠E，∴∠E = ∠C，∴DC = DE = 4. ∵∠ADB = 90°，∴∠ADC = 90°，∴AD = $\sqrt{AC^{2}-CD^{2}}$ = $\sqrt{5^{2}-4^{2}}$ =
3
.
答案：解析 （1）證明：∵AB是⊙O的直徑，∴∠ADB = 90°，∴AD⊥BD，又∵AB = AC，∴BD = CD.
（2）∵AB = 5，∴AC = AB = 5，∴∠B = ∠C. ∵∠B = ∠E，∴∠E = ∠C，∴DC = DE = 4. ∵∠ADB = 90°，∴∠ADC = 90°，∴AD = $\sqrt{AC^{2}-CD^{2}}$ = $\sqrt{5^{2}-4^{2}}$ = 3.
15.新 課標 推理能力「2025湖北荊州期中」如圖,AB為⊙O的直徑,點C,D為直徑AB同側(cè)圓上的點,且點D為$\overset{\frown}{AC}$的中點,過點D作DE⊥AB于點E,交AC于點G,延長DE,交⊙O于點F.
(1)如圖①,若∠BAC= 30°,求證：$\overset{\frown}{CD}= \overset{\frown}{BC}$.
(2)如圖②,若AC= 12,BE= 9,求⊙O的半徑.
答案：解析 （1）證明：如圖①，連接OC，OD，∵∠BAC = 30°，∴∠BOC = 2∠BAC = 60°，∴∠AOC = 180° - 60° = 120°，∵點D為$\overset{\frown}{AC}$的中點，∴$\overset{\frown}{AD}$ = $\overset{\frown}{DC}$，∴∠AOD = ∠COD = 60°，∴∠COD = ∠COB，∴$\overset{\frown}{CD}$ = $\overset{\frown}{BC}$.
（2）如圖②，連接OF，∵DE⊥AB，AB為⊙O的直徑，∴$\overset{\frown}{AD}$ = $\overset{\frown}{AF}$，DE = EF，∵$\overset{\frown}{AD}$ = $\overset{\frown}{DC}$，∴$\overset{\frown}{AD}$ + $\overset{\frown}{DC}$ = $\overset{\frown}{AD}$ + $\overset{\frown}{AF}$，∴$\overset{\frown}{AC}$ = $\overset{\frown}{DF}$，∴DF = AC = 12，∴EF = 6，設(shè)⊙O的半徑為r，則OE = BE - OB = 9 - r，在Rt△EOF中，EO2 + EF2 = OF2，即(9 - r)2 + 62 = r2，解得r = 6.5，∴⊙O的半徑為6.5.
