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17. (10分)(2026·南充期末)如图,AB,ADC分别是半圆O的直径和割线,弦BE平分∠ABD,OE与AC交于F,EG⊥AB于G,∠OEG=∠DBC.
(1)求证:BC是半圆O的切线;
(2)若$BE=4\sqrt{5}$,$EF=2$,求EG的长.
(1)求证:BC是半圆O的切线;
(2)若$BE=4\sqrt{5}$,$EF=2$,求EG的长.
答案
17.(1)$\because AB$是半圆O的直径,$\therefore ∠ ADB=90°$.$\because$ 弦BE平分$∠ ABD$,$\therefore ∠ ABE=∠ EBD$,$\therefore \widehat{AE}=\widehat{ED}$,$\therefore OE⊥ AD$,$\therefore ∠ AFO=90°$,$\therefore OF// BD$,$\therefore ∠ AOF=∠ ABD$,$\therefore ∠ OEG=90°-∠ AOF=90°-∠ ABD=∠ BAD$.$\because ∠ OEG=∠ DBC$,$\therefore ∠ ABC=∠ ABD+∠ DBC=∠ ABD+∠ BAD=90°$,$\therefore BC$是半圆O的切线.
(2)设半圆O的半径为x,$\therefore OF=OE-EF=x-2$.$\because OE=OA$,$∠ GOE=∠ FOA$,$∠ EGO=∠ AFO=90°$,$\therefore △ GOE≌△ FOA$,$\therefore OG=OF=x-2$,$BG=OB+OG=2x-2$,由勾股定理得$EG^2=BE^2-BG^2=OE^2-OG^2$,$\therefore (4\sqrt{5})^2-(2x-2)^2=x^2-(x-2)^2$,解得$x_1=-4$(舍去),$x_2=5$,$\therefore EG=\sqrt{OE^2-OG^2}=\sqrt{5^2-3^2}=4$,$\therefore EG$的长为4.
(2)设半圆O的半径为x,$\therefore OF=OE-EF=x-2$.$\because OE=OA$,$∠ GOE=∠ FOA$,$∠ EGO=∠ AFO=90°$,$\therefore △ GOE≌△ FOA$,$\therefore OG=OF=x-2$,$BG=OB+OG=2x-2$,由勾股定理得$EG^2=BE^2-BG^2=OE^2-OG^2$,$\therefore (4\sqrt{5})^2-(2x-2)^2=x^2-(x-2)^2$,解得$x_1=-4$(舍去),$x_2=5$,$\therefore EG=\sqrt{OE^2-OG^2}=\sqrt{5^2-3^2}=4$,$\therefore EG$的长为4.
18. (10分)(2026·安阳月考)如图,六边形ABCDEF是$\odot O$的内接正六边形,四边形EFGH是正方形,连接OE,OF,OG.
(1)求$∠ OGF$的度数;
(2)取劣弧$\overset{\frown}{CD}$的中点K,连接FK,在图中找出和FK等长的线段,并说明理由.
(1)求$∠ OGF$的度数;
(2)取劣弧$\overset{\frown}{CD}$的中点K,连接FK,在图中找出和FK等长的线段,并说明理由.
答案
18.(1)$\because ∠ FOE$为正六边形的中心角,$\therefore ∠ FOE=60°$.$\because EO=FO$,$\therefore △ EOF$是等边三角形,$\therefore ∠ OFE=60°$,$OF=EF$.又四边形EFGH是正方形,$\therefore OF=EF=FG$,$\therefore △ OFG$是等腰三角形.$\because ∠ EFG=90°$,$∠ OFE=60°$,$\therefore ∠ OFG=150°$,$\therefore ∠ OGF=\frac{1}{2}×(180°-150°)=15°$.
(2)GO与FK是等长的线段,理由:如图,连接OK,$\because K$是$\widehat{CD}$的中点,$\therefore OK$垂直平分CD,在正六边形ABCDEF中,$∠ CDE=120°$,$∠ OED=60°$,$\therefore ∠ CDE+∠ OED=180°$,$\therefore OE// CD$,$\therefore ∠ EOK=90°$,$\therefore ∠ FOK=60°+90°=150°$,在$△ OFK$和$△ FGO$中,$\begin{cases}OF=FG,\\∠ FOK=∠ GFO,\\OK=FO,\end{cases}$$\therefore △ OFK≌△ FGO(\mathrm{SAS})$,$\therefore FK=GO$.
19. (12分)(2026·昆明月考)如图,四边形ABCD内接于⊙O,对角线AC与BD交于点E,AC平分∠BAD,∠BCF=∠CAB.
(1)若∠ABC=110°,求∠ADC的度数.
(2)求证:CF是⊙O的切线.
(3)已知∠BAD=60°,是否存在常数a,b使等式aAB+bAD=√3 AC成立?若存在,请直接写出一个a的值和一个b的值,并证明你写出的a的值和b的值,使等式aAB+bAD=√3 AC成立;若不存在,请说明理由.
备用图
(1)若∠ABC=110°,求∠ADC的度数.
(2)求证:CF是⊙O的切线.
(3)已知∠BAD=60°,是否存在常数a,b使等式aAB+bAD=√3 AC成立?若存在,请直接写出一个a的值和一个b的值,并证明你写出的a的值和b的值,使等式aAB+bAD=√3 AC成立;若不存在,请说明理由.
备用图
答案
19.(1)$\because$ 四边形ABCD是$\odot O$的内接四边形,$\therefore ∠ ABC+∠ ADC=180°$.又$\because ∠ ABC=110°$,$\therefore ∠ ADC=180°-110°=70°$.
(2)如图①,连接OC,OB,$\because ∠ CAB=\frac{1}{2}∠ BOC$,$∠ BCF=∠ CAB$,$\therefore ∠ BCF=\frac{1}{2}∠ BOC$.又$\because BO=CO$,$\therefore ∠ OBC=∠ BCO$,在$△ BOC$中,$∠ OBC+∠ BCO+∠ BOC=180°$,$\therefore ∠ BCO=90°-\frac{1}{2}∠ BOC$,$\therefore ∠ OCF=∠ BCO+∠ BCF=90°-\frac{1}{2}∠ BOC+\frac{1}{2}∠ BOC=90°$,$\therefore OC⊥ CF$.又$\because OC$为$\odot O$的半径,$\therefore CF$是$\odot O$的切线.
(3)存在常数a,b,使$aAB+bAD=\sqrt{3}AC$成立,且a=1,b=1,证明如下:
如图②,延长AD至G,使DG=AB,过点C作$CH⊥ AD$于点H,连接CG,$\because AC$平分$∠ BAD$,$∠ BAD=60°$,$\therefore ∠ BAC=∠ CAD=30°$,$\therefore \widehat{BC}=\widehat{CD}$,$\therefore BC=CD$.$\because$ 四边形ABCD是$\odot O$的内接四边形,$\therefore ∠ ABC+∠ ADC=180°$.又$\because ∠ ADC+∠ CDG=180°$,$\therefore ∠ ABC=∠ CDG$,在$△ ABC$和$△ GDC$中,$\begin{cases}BC=DC,\\∠ ABC=∠ GDC,\\AB=GD,\end{cases}$$\therefore △ ABC≌△ GDC(\mathrm{SAS})$,$\therefore AC=GC$.$\because CH⊥ AD$,$\therefore AH=GH=\frac{1}{2}AG$,$∠ AHC=90°$,$\therefore DG+AD=AB+AD=AG=2AH$,在$\mathrm{Rt}△ ACH$中,$∠ AHC=90°$,$∠ CAH=30°$,$AH^2+CH^2=AC^2$,$\therefore CH=\frac{1}{2}AC$,$\therefore AH^2+(\frac{1}{2}AC)^2=AC^2$,$\therefore AH=\frac{\sqrt{3}}{2}AC$,$\therefore AB+AD=AG=2×\frac{\sqrt{3}}{2}AC=\sqrt{3}AC$,$\therefore$ 当a=1,b=1时,$aAB+bAD=\sqrt{3}AC$成立.
