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8. (教材$P_{123}T_{2}$变式)如图,在$\odot O$中,C 为$\widehat {AB}$的中点,$CD⊥OA$于点 D,$CE⊥OB$于点 E. 求证:
$OD= OE.$
$OD= OE.$
答案
证明:连接 $OC$. $\because C$ 为 $\widehat{AB}$ 的中点,
$\therefore \widehat{AC}=\widehat{BC},\therefore ∠AOC = ∠BOC$.
$\because CD⊥OA,CE⊥OB,\therefore CD = CE$.
$\because OC = OC$,
$\therefore Rt△OCD≌Rt△OCE(HL)$,
$\therefore OD = OE$.
$\therefore \widehat{AC}=\widehat{BC},\therefore ∠AOC = ∠BOC$.
$\because CD⊥OA,CE⊥OB,\therefore CD = CE$.
$\because OC = OC$,
$\therefore Rt△OCD≌Rt△OCE(HL)$,
$\therefore OD = OE$.
9. 如图,AB 是$\odot O$的直径,M,N 分别是 AO,BO 的中点,$CM⊥AB,DN⊥AB$. 求证:$\widehat {AC}= \widehat {BD}.$
答案
证明:连接 $OC,OD$.
$\because M,N$ 分别是 $AO,BO$ 的中点,
$\therefore OM=\frac{1}{2}AO,ON=\frac{1}{2}BO$.
$\because AO = BO,\therefore OM = ON$.
$\because CM⊥AB,DN⊥AB$,
$\therefore ∠CMO = ∠DNO = 90^{\circ}$.
在 $Rt△COM$ 和 $Rt△DON$ 中,
$\left\{\begin{array}{l} OC = OD,\\ OM = ON,\end{array}\right.$
$\therefore Rt△COM≌Rt△DON$,
$\therefore ∠COM = ∠DON$,
$\therefore \widehat{AC}=\widehat{BD}$.
$\because M,N$ 分别是 $AO,BO$ 的中点,
$\therefore OM=\frac{1}{2}AO,ON=\frac{1}{2}BO$.
$\because AO = BO,\therefore OM = ON$.
$\because CM⊥AB,DN⊥AB$,
$\therefore ∠CMO = ∠DNO = 90^{\circ}$.
在 $Rt△COM$ 和 $Rt△DON$ 中,
$\left\{\begin{array}{l} OC = OD,\\ OM = ON,\end{array}\right.$
$\therefore Rt△COM≌Rt△DON$,
$\therefore ∠COM = ∠DON$,
$\therefore \widehat{AC}=\widehat{BD}$.
10. 如图,AB 为$\odot O$的直径,C 是圆上一点,D 是$\widehat {BC}$的中点,弦$DE⊥AB$,垂足为 F.
(1)求证:$BC= 2DF;$
(2)若$AC= 6,BF= 2$,求 BC 的长.
(1)求证:$BC= 2DF;$
(2)若$AC= 6,BF= 2$,求 BC 的长.
答案
解:(1)$\because D$ 是 $\widehat{BC}$ 的中点,
$\therefore \widehat{CD}=\widehat{BD}$.
$\because DE⊥AB$,且 $AB$ 为 $\odot O$ 的直径,
$\therefore \widehat{BE}=\widehat{BD}$,
$\therefore \widehat{BC}=\widehat{DE},\therefore BC = DE$.
$\because DE⊥AB,\therefore DE = 2DF$,
$\therefore BC = DE = 2DF$;
(2)连接 $OD$,交 $BC$ 于点 $H$.
$\because \widehat{CD}=\widehat{BD},\therefore DO⊥BC$,
$\therefore ∠BHO = 90^{\circ},BH = CH$,
$\because DE⊥AB,\therefore ∠DFO = 90^{\circ}$,
$\therefore ∠BHO = ∠DFO$.
$\because ∠DOF = ∠BOH$,
$OB = OD$,
$\therefore △BOH≌△DOF$,
$\therefore OF = OH=\frac{1}{2}AC = 3$,
$\therefore OB = OF + FB = 5,\therefore OD = 5$.
在 $Rt△ODF$ 中,
$DF=\sqrt{OD^{2}-OF^{2}} = 4$,
$\therefore BC = DE = 2DF = 8$.
$\therefore \widehat{CD}=\widehat{BD}$.
$\because DE⊥AB$,且 $AB$ 为 $\odot O$ 的直径,
$\therefore \widehat{BE}=\widehat{BD}$,
$\therefore \widehat{BC}=\widehat{DE},\therefore BC = DE$.
$\because DE⊥AB,\therefore DE = 2DF$,
$\therefore BC = DE = 2DF$;
(2)连接 $OD$,交 $BC$ 于点 $H$.
$\because \widehat{CD}=\widehat{BD},\therefore DO⊥BC$,
$\therefore ∠BHO = 90^{\circ},BH = CH$,
$\because DE⊥AB,\therefore ∠DFO = 90^{\circ}$,
$\therefore ∠BHO = ∠DFO$.
$\because ∠DOF = ∠BOH$,
$OB = OD$,
$\therefore △BOH≌△DOF$,
$\therefore OF = OH=\frac{1}{2}AC = 3$,
$\therefore OB = OF + FB = 5,\therefore OD = 5$.
在 $Rt△ODF$ 中,
$DF=\sqrt{OD^{2}-OF^{2}} = 4$,
$\therefore BC = DE = 2DF = 8$.
11. (原创题)如图,点 A,B,C,D 都在$\odot O$上,$∠AOB+∠COD= 90^{\circ }.$
(1)将$△COD$绕点 O 逆时针旋转得到$△BOE$,使 OC 与 OB 重合,画出旋转后的图形;
(2)在(1)的条件下,求$∠ABE$的度数;
(3)在(1)的条件下,若$AB= 2,CD= \sqrt {2}$,求$\odot O$的半径.
(1)将$△COD$绕点 O 逆时针旋转得到$△BOE$,使 OC 与 OB 重合,画出旋转后的图形;
(2)在(1)的条件下,求$∠ABE$的度数;
(3)在(1)的条件下,若$AB= 2,CD= \sqrt {2}$,求$\odot O$的半径.
答案
解:(1)过点 $O$ 作 $OE⊥OA$,
交 $\odot O$ 于点 $E$,
连接 $BE$,则 $△OBE$ 即为所求;
(2)$\because ∠AOB + ∠COD = 90^{\circ}$,
$\therefore ∠AOE = 90^{\circ}$.
$\because OA = OB = OE$,
$\therefore ∠OAB = ∠OBA$,
$∠OBE = ∠OEB$.
$\because ∠OAB + ∠OBA + ∠OBE + ∠OEB + ∠AOE = 360^{\circ}$,
$\therefore ∠ABE = (360^{\circ}-90^{\circ})÷2 = 135^{\circ}$;
(3)连接 $AE$,过点 $E$ 作 $EF⊥AB$ 交 $AB$ 的延长线于点 $F$,
$\therefore ∠EBF = 45^{\circ}$.
$\because BE = CD=\sqrt{2}$,
$\therefore EF = BF = 1$,
$\therefore AF = AB + BF = 3$,
$\therefore AE=\sqrt{3^{2}+1^{2}}=\sqrt{10}$.
$\because OA = OE,∠AOE = 90^{\circ}$,
$\therefore OA=\frac{\sqrt{2}}{2}AE=\frac{\sqrt{2}}{2}×\sqrt{10}=\sqrt{5}$.
