7 (2024 连云港东海期中)如图，A，B，C，D 是$\odot O$上的四个点，$AD=BC$. 求证：$AB=CD$.

证明：∵​$AD = BC，$​∴​$\widehat {AD}=\widehat {BC}$​∴​$\widehat {AD}+\widehat {AC}=\widehat {BC}+\widehat {AC}，$​即​$\widehat {CD}=\widehat {AB}$​∴​$AB = CD$​

8 如图,在$\odot O$中,弦$AB$,$CD$相交于点$E$,且$AB=CD$.求证:$AC=BD$.

证明：∵​$AB = CD，$​∴​$\widehat {AB}=\widehat {CD}$​∴​$\widehat {AC}=\widehat {BD}$​∴​$AC = BD$​

9 如图,在$\odot O$中,$\overset{\frown}{AC}=\overset{\frown}{CB}$,$CD⊥ AO$于点$D$,$CE⊥ OB$于点$E$.
(1) 求证:$AD=BE$;
(2) 若$AD=DO$,$r=3$,求$CD$的长.

​$ (1)$​证明：连接​$OC$​∵​$\widehat {AC}=\widehat {BC}，$​∴​$∠AOC=∠BOC$​又​$CD\perp OA，$​​$CE\perp OB，$​∴​$CD = CE$​在​$\triangle COD$​和​$\triangle COE$​中​$\begin {cases}∠COD=∠COE\\∠CDO=∠CEO\\OC = OC\end {cases}$​∴​$\triangle COD≌\triangle COE(\mathrm {AAS})，$​∴​$OD = OE$​∵​$OA = OB，$​∴​$AD = BE$​​$ (2)$​解：∵​$AD = DO，$​​$r = 3，$​∴​$AD = DO=\frac 32$​又∵​$CD\perp OA$​∴在​$Rt\triangle CDO$​中，​$CD=\sqrt {CO^2-DO^2}=\sqrt {3^2-(\frac 32)^2}=\frac {3\sqrt 3}2$​ ;

10 如图,AC,BD是$\odot O$的弦,分别连接$OA,OB,OC,OD,∠ AOB=∠ COD$,AC与OB交于点E,$\odot O$的半径为6.
(1) 求证:$\overset{\frown}{AC}=\overset{\frown}{BD}$;
(2) 若$BD=10$,E为AC的中点,求OE的长.

​$ (1)$​证明：∵​$∠AOB=∠COD，$​∴​$\widehat {AB}=\widehat {CD}$​又∵​$\widehat {AC}=\widehat {AB}+\widehat {BC}，$​​$\widehat {BD}=\widehat {CD}+\widehat {BC}$​∴​$\widehat {AC}=\widehat {BD}$​​$ (2)$​解：∵​$\widehat {AC}=\widehat {BD}，$​∴​$AC = BD = 10$​∵​$OA = OC，$​​$E$​为​$AC$​的中点，∴​$OE\perp AC$​∴​$AE=\frac 12\ \mathrm {A}C = 5$​在​$Rt\triangle AEO$​中，由勾股定理，得​$OE=\sqrt {OA^2-AE^2}=\sqrt {6^2-5^2}=\sqrt {11}$​∴​$OE$​的长是​$\sqrt {11}$​