Die Winkelsumme im Dreieck beträgt $180^\circ$

$\def\arraystretch{1.25} \begin{array}{l}\alpha+\beta+100^\circ=180^\circ\\\alpha+\beta=80^\circ\end{array}$

Die Summe von Nebenwinkeln beträgt $180^\circ$.

$\def\arraystretch{1.25} \begin{array}{l}\alpha+\alpha\ast=180^\circ\\\alpha=180^\circ-\alpha\ast\\\\\beta+\beta\ast=180^\circ\\\beta=180^\circ-\beta\ast\end{array}$

$\def\arraystretch{1.25} \begin{array}{l}(180^\circ-\alpha\ast)+(180^\circ-\beta\ast)=80^\circ\end{array}$

$\def\arraystretch{1.25} \begin{array}{l}360^\circ-\alpha\ast-\beta\ast=80^\circ\\\\\end{array}$

$280^\circ=\alpha\ast+\beta\ast$

Die Lösung lautet $\alpha\ast+\beta\ast=280^\circ$