7. O是$△ ABC$三条边的垂直平分线的交点，$OA=8$，则$OA+OB+OC$的值是。

7.24

8. 已知 $C,D$ 两点在线段 $AB$ 的垂直平分线上, 且 $∠ ACB=50°,∠ ADB=90°$, 则 $∠ CAD=$。

8.110°或20°

9. (2024·淮安期末) 如图,在$△ ABC$中,$AD ⊥ BC$,$EF$垂直平分$AC$,交$AC$于点$F$,交$BC$于点$E$,且$BD=DE$.
(1)若$∠ BAE=40°$,求$∠ C$的度数;
(2)若$△ ABC$的周长为$20\ \mathrm{cm}$,$AC=8\ \mathrm{cm}$,求$DC$长.

9.解:(1)$\because AD⊥ BC$,$BD=DE$,
$\therefore AD$垂直平分线段$BE$,$\therefore AB=AE$,$\therefore ∠ B=∠ AEB$.
$\because ∠ BAE=40°$,$\therefore ∠ AEB=\dfrac{1}{2}(180°-∠ BAE)=70°$.
$\because EF$垂直平分$AC$,$\therefore EA=EC$,$\therefore ∠ C=∠ EAC$.
又$\because ∠ AEB=∠ EAC+∠ C$,$\therefore ∠ C=\dfrac{1}{2}∠ AEB=35°$.
(2)$\because △ ABC$的周长为$20\ \mathrm{cm}$,$AC=8\ \mathrm{cm}$,
$\therefore AB+BC=12\ \mathrm{cm}$.
$\because AB=AE=CE$,$BD=DE$,
$\therefore CE+DE=\dfrac{1}{2}(AB+BC)=6\ \mathrm{cm}$,即$DC=6\ \mathrm{cm}$.

10. (2024·海州区期中) 如图,在$△ ABC$中,$AC$边的垂直平分线分别交$BC$,$AC$于点$E$,$F$,连接$AE$,作$AD⊥ BC$于点$D$,且$D$为$BE$的中点.
(1)试说明:$AB=CE$;
(2)若$∠ C=32°$,求$∠ BAC$的度数.

10.解:(1)$\because D$为$BE$的中点,$\therefore BD=DE$.
又$\because AD⊥ BC$于点$D$,$\therefore AB=AE$.
$\because EF$是$AC$的垂直平分线,$\therefore AE=EC$,$\therefore AB=CE$.
(2)$\because ∠ C=32°$,$AE=EC$,$\therefore ∠ EAC=∠ C=32°$,
$\therefore ∠ AEB=∠ EAC+∠ C=32°+32°=64°$.
又$\because AB=AE$,$\therefore ∠ B=∠ AEB=64°$,
$\therefore ∠ BAE=180°-∠ B-∠ AEB=52°$,
$\therefore ∠ BAC=∠ BAE+∠ CAE=52°+32°=84°$.

11. (2024·南充) 如图,在$△ ABC$中,$D$为$BC$边的中点,过点$B$作$BE // AC$交$AD$的延长线于点$E$.
(1)求证:$△ BDE ≌ △ CDA$;
(2)若$AD ⊥ BC$,求证:$BA=BE$.

11.证明:(1)$\because D$为$BC$边的中点,$\therefore BD=CD$.
$\because BE// AC$,$\therefore ∠ DBE=∠ C$,$∠ E=∠ DAC$.
在$△ BDE$和$△ CDA$中,$\begin{cases}∠ E=∠ DAC,\\∠ EBD=∠ C,\\BD=CD,\end{cases}$
$\therefore △ BDE≌ △ CDA(\mathrm{AAS})$.
(2)$\because △ BDE≌ △ CDA$,$\therefore ED=AD$.
又$\because AD⊥ BC$,$\therefore BC$垂直平分线段$AE$,$\therefore BA=BE$.