用两种方法证明“三角形的外角和等于 $360°$”．

如图， $\angle \mathit{BAE}$、 $\angle \mathit{CBF}$、 $\angle \mathit{ACD}$是 $\mathit{\Delta ABC}$的三个外角．

求证 $\angle \mathit{BAE}+\angle \mathit{CBF}+\angle \mathit{ACD}=360°$．

证法 $1:\because$　　，

$\therefore \angle \mathit{BAE}+\angle 1+\angle \mathit{CBF}+\angle 2+\angle \mathit{ACD}+\angle 3=180°\times 3=540°$

$\therefore \angle \mathit{BAE}+\angle \mathit{CBF}+\angle \mathit{ACD}=540°-(\angle 1+\angle 2+\angle 3)$．

$\because$　　，

$\therefore \angle \mathit{BAE}+\angle \mathit{CBF}+\angle \mathit{ACD}=540°-180°=360°$．

请把证法1补充完整，并用不同的方法完成证法2．