《2025年启东中学作业本九年级数学上册苏科版连淮专版》

8. 如图,在扇形OAB中,$∠AOB= 110^{\circ }$,将扇形OAB沿过点B的直线折叠,点O恰好落在$\widehat {AB}$上的点D处,折痕交OA于点C,则$\widehat {AD}$的度数为 (B)
A.$40^{\circ }$
B.$50^{\circ }$
C.$60^{\circ }$
D.$70^{\circ }$

答案： B

9. 如图,C是$\odot O$的直径AB上的一点,过点C作弦DE,使$CD= CO$.若$\widehat {AD}的度数为35^{\circ }$,则$\widehat {BE}$的度数是105°.

答案： 105°

10. 如图,以平行四边形ABCD的顶点A为圆心,AB长为半径作$\odot A$,分别交BC,AD于E,F两点,交BA的延长线于点G.
(1)求证:$\widehat {EF}= \widehat {FG};$
(2)连接AE,若$∠EAG= 140^{\circ }$,求$∠D$的度数.

答案：

(1)证明:如答图,连接AE.
∵四边形ABCD是平行四边形,
∴AD // BC,
∴∠EAF = ∠AEB,∠GAF = ∠B.
∵AE = AB,
∴∠B = ∠AEB,
∴∠EAF = ∠GAF,
∴$\overset{\frown}{EF} = \overset{\frown}{FG}$.
(2)解:
∵∠EAG = 140°,
∴∠BAE = 180° - 140° = 40°,
∴∠AEB = ∠ABE = $\frac{1}{2}$×(180° - 40°) = 70°.
∵四边形ABCD为平行四边形,
∴∠D = ∠ABE = 70°.

11. 如图,在$\odot O$中,C,D是直径AB上两点,且$AC= BD,MC⊥AB,ND⊥AB$,垂足分别为C,D,点M,N在$\odot O$上.
(1)求证:$\widehat {AM}= \widehat {BN};$
(2)若C,D分别为OA,OB的中点,则$\widehat {AM}= \widehat {MN}= \widehat {NB}$成立吗? 请说明理由.

答案：

(1)证明:连接OM,ON,如答图.
∵AC = BD,
∴OA - AC = OB - BD,即OC = OD.
∵MC⊥AB,ND⊥AB,
∴∠OCM = 90°,∠ODN = 90°.在Rt△OCM和Rt△ODN中,$\begin{cases} OM = ON, \\ OC = OD, \end{cases}$
∴Rt△OCM ≌ Rt△ODN (HL),
∴∠COM = ∠DON,
∴$\overset{\frown}{AM} = \overset{\frown}{BN}$.
(2)解:$\overset{\frown}{AM} = \overset{\frown}{MN} = \overset{\frown}{NB}$成立.理由:
∵C,D分别为OA,OB的中点,
∴OC = $\frac{1}{2}$OA,OD = $\frac{1}{2}$OB,
∴OC = $\frac{1}{2}$OM,OD = $\frac{1}{2}$ON,
∴∠OMC = 30°,∠OND = 30°,
∴∠MOC = ∠NOD = 60°,
∴∠MON = 60°,
∴∠AOM = ∠MON = ∠BON,
∴$\overset{\frown}{AM} = \overset{\frown}{MN} = \overset{\frown}{NB}$.