$\widehat{BOC}=180^{\circ}-(\widehat{OBC}+\widehat{OCB})=180^{\circ}-(\frac{\widehat{ABC}}{2}+\frac{\widehat{ACB}}{2})$
Mà $\widehat{BOC}=120^{\circ}$ nên $\widehat{ABC}+\widehat{ACB}=120$ (1)
Do $\Delta ABC=\Delta MNP$ nên:
$\widehat{ABC}=\widehat{MNP}$ và $\widehat{ACB}=\widehat{MPN}$
hay $\widehat{ABC}+\widehat{ACB}=\widehat{MNP}+\widehat{MPN}$ (2)
Từ (1) và 92), ta có $\widehat{MNP}+\widehat{MPN}=120^{\circ}$