13. 如图，AB是$\odot O$的直径，C为$\overset{\frown}{AE}$的中点，$CD⊥ AB$于点D，交AE于点F，连接AC。求证：$AF=CF$。

13. 证明:连接$BC$,$\because AB$是$\odot O$的直径,$\therefore ∠ACB=90°$,即$∠ACF+∠BCD=90°$,$\because CD⊥AB$,$\therefore ∠B+∠BCD=90°$,$\therefore ∠ACF=∠B$,$\because C$为$\overset{\frown}{AC}$的中点,$\therefore \overset{\frown}{AC}=\overset{\frown}{CE}$. $\therefore ∠B=∠CAE$,$\therefore ∠ACF=∠CAE$,$\therefore AF=CF$.

14. 如图，AB是$\odot O$的直径，点D，E在$\odot O$上，AE，BD的延长线交于点C，且AB=AC. 求证：BD=DE.

14. 证明:连接$AD$. $\because AB$为$\odot O$的直径,点$D$在$\odot O$上,$\therefore ∠ADB=90°$,即$AD⊥BC$. 又$\because AB=AC$,$\therefore ∠BAD=∠DAE$,则$\overset{\frown}{BD}=\overset{\frown}{DE}$. $\therefore BD=DE$.

15.(核心素养·推理能力)如图,在$\odot O$中,弦$AC⊥ BD$于点$E$,连接$AB,CD,BC,OA,OB,OC,OD$.
(1)求证:$∠ AOB+∠ COD=180°$;
(2)若$AB=8,CD=6$,求$\odot O$的直径.

15. (1)证明:由圆周角定理,得$∠AOB=2∠ACB$,$∠COD=2∠CBD$. $\because AC ⊥ BD$,$\therefore ∠BEC = 90°$,$\therefore ∠ACB+∠CBD = 90°$,$\therefore ∠AOB+∠COD = 180°$.
(2)解:延长$BO$交$\odot O$于点$F$,连接$AF,DF,AD$. $\because BF$是$\odot O$的直径,$\therefore ∠BDF=90°$.即$DF⊥BD$. $\because AC⊥BD$,$\therefore AC // DF$,$\therefore ∠CAD=∠ADF$,$\therefore AF=CD=6$. $\because BF$是$\odot O$的直径,$\therefore ∠BAF=90°$,$\therefore$在$\mathrm{Rt}△ABF$中,$BF = \sqrt{AB^2+AF^2}=\sqrt{8^2+6^2}=10$,$\therefore \odot O$的直径为10.

变式1 如图，$△ ABC$内接于$\odot O$，$AB=10\mathrm{cm}$，$AC=6\mathrm{cm}$，弦$CD$平分$∠ ACB$，$∠ ACB=90°$，则$CD=\_\_\_\_\_\_\mathrm{cm}$.

变式1:$7\sqrt{2}$

变式2 连接上图中的AD，BD，则四边形ACBD的面积为$\mathrm{cm}^2$。

变式2:$49$

拓展1 连接上图中的AD,把变式1中的条件$AB=10\mathrm{cm}$,改为$AD=5\sqrt{2}\mathrm{cm}$,其他条件不变,则$BC=\_\_\_\_\_\_\mathrm{cm}$.

拓展1:$8$

拓展2 若变式1中的条件$∠ ACB=90°$改为$∠ ACB=120°$，其他条件不变，则$CD=$ cm.

拓展2:$3+\sqrt{73}$