《2025年学霸提优大试卷九年级数学全一册苏科版》

23. (10分)(临沂中考)如图,AB是$\odot O$的切线,B为切点,直线AO交$\odot O$于C、D两点,连接BC、BD.过圆心O作BC的平行线,分别交AB的延长线、$\odot O$及BD于点E、F、G.
(1)求证:$∠D= ∠E$;
(2)若F是OE的中点,$\odot O$的半径为3,求阴影部分的面积.

答案：
(1) 如图，连接 OB.$\because AB$是$\odot O$的切线，$\therefore \angle OBE=90^{\circ }$，$\therefore \angle E+\angle BOE=90^{\circ }$.$\because CD$为$\odot O$的直径，$\therefore \angle CBD=90^{\circ }$，$\therefore \angle D+\angle DCB=90^{\circ }$.$\because OE// BC$，$\therefore \angle BOE=\angle OBC$.$\because OB=OC$，$\therefore \angle OBC=\angle OCB$，$\therefore \angle BOE=\angle OCB$，$\therefore \angle D=\angle E$.
(2) $\because F$为 OE 的中点，$OB=OF$，$\therefore OF=EF=3$，$\therefore OE=6$，$\therefore BO=\frac{1}{2}OE$.$\because \angle OBE=90^{\circ }$，$\therefore \angle E=30^{\circ }$，$\therefore \angle BOG=60^{\circ }$.$\because OE// BC$，$\angle DBC=90^{\circ }$，$\therefore \angle OGB=90^{\circ }$，$\therefore OG=\frac{3}{2}$，$BG=\frac{3}{2}\sqrt{3}$，$\therefore S_{\triangle BOG}=\frac{1}{2}OG\cdot BG=\frac{1}{2}× \frac{3}{2}× \frac{3}{2}\sqrt{3}=\frac{9}{8}\sqrt{3}$，$S_{\text{扇形}BOF}=\frac{60× \pi × 3^{2}}{360}=\frac{3}{2}\pi$，$\therefore S_{\text{阴影}}=S_{\text{扇形}BOF}-S_{\triangle BOG}=\frac{3}{2}\pi -\frac{9}{8}\sqrt{3}$.

24. (12分)如图①,四边形ADBC内接于$\odot O$,E为BD延长线上一点,DA平分$∠EDC$.
(1)求证:$AB= AC$;
(2)如图②,CD为直径,过点A的圆的切线交BD的延长线于点E,若$DE= 1,AE= 2$,求$\odot O$的半径.

答案：
(1) $\because$四边形 ADBC 内接于$\odot O$，$\therefore \angle EDA=\angle ACB$.由圆周角定理得$\angle CDA=\angle ABC$.$\because DA$平分$\angle EDC$，$\therefore \angle EDA=\angle CDA$，$\therefore \angle ABC=\angle ACB$，$\therefore AB=AC$.
(2) 如图，连接 AO 并延长交 BC 于点 H，作 AM⊥CD 于点 M.$\because AB=AC$，$\therefore AH\perp BC$.又$AH\perp AE$，$\therefore AE// BC$.$\because CD$为$\odot O$的直径，$\therefore \angle DBC=90^{\circ }$，$\therefore \angle E=\angle DBC=90^{\circ }$，$\therefore$四边形 AEBH 为矩形，$\therefore BH=AE=2$，$\therefore BC=4$.$\because DA$平分$\angle EDC$，$\angle E=90^{\circ }$，$AM\perp CD$，$\therefore DE=DM=1$，$AE=AM=2$.在$\text{Rt}\triangle ABE$和$\text{Rt}\triangle ACM$中，$\begin{cases}AE=AM\\AB=AC\end{cases}$，$\therefore \text{Rt}\triangle ABE≌\text{Rt}\triangle ACM(HL)$，$\therefore BE=CM$.设 BE=x，则 BD=x-1，$CD=x+1$.在$\text{Rt}\triangle BDC$中，$(x - 1)^{2}+4^{2}=(x + 1)^{2}$，解得 x=4，$\therefore CD=5$，$\therefore \odot O$的半径为 2.5.