We have:

\(\widehat{A_1}+\widehat{D_2}+\widehat{C}=180^o\)

\(\widehat{A_2}+\widehat{D_1}+\widehat{B}=180^o\)

\(\Rightarrow\widehat{A_1}+\widehat{D_2}+\widehat{C}+\widehat{A_2}+\widehat{D_1}+\widehat{B}=180^0+180^0\)

=> \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

So the sum of four angles in a quadrilateral is 360^o

Connect A and C 

Following total 3 corners in a triangle, we have :

\(\widehat{ADC}+\widehat{DCA}+\widehat{CAD}=180^0\)  (\(\Delta ACD\))

\(\widehat{ACB}+\widehat{CBA}+\widehat{BAC}=180^0\)  (\(\Delta ABC\))

=> \(\widehat{ADC}+\widehat{DCA}+\widehat{CAD}\) + \(\widehat{ACB}+\widehat{CBA}+\widehat{BAC}\) = 180⁰ + 180⁰ = 360⁰

<=> \(\widehat{ADC}+\widehat{DCB}+\widehat{CBA}+\widehat{BAD}=360^0\)