Nối đoạn thẳng BD.
Xét tam giác ABD có: $\widehat{A}+\widehat{ABD}+\widehat{ADB}=180^{o}$.
Xét tam giác BDC có: $\widehat{C}+\widehat{CBD}+\widehat{CDB}=180^{o}$.
$\Rightarrow $ $\widehat{A} + \widehat{B}+\widehat{C}+\widehat{D}$ = $\widehat{A}+\widehat{ABD}+\widehat{CBD}+\widehat{ADB} + \widehat{CDB} +\widehat{C}= 180^{o} + 180^{o} = 360^{o}$.
Vậy $\widehat{A} + \widehat{B}+\widehat{C}+\widehat{D} = 360^{o}$