Ta có:
$\widehat{AQB} = 180^o - (\widehat{QAB} + \widehat{QBA})$
$=180^o - \dfrac{\widehat{A} + \widehat{B}}{2}$
Tương tự, ta được:
$\widehat{DPC} = 180^o - \dfrac{\widehat{D} + \widehat{C}}{2}$
$\Rightarrow \widehat{AQB} + \widehat{DPC} = 180^o- \dfrac{\widehat{A} + \widehat{B}}{2} + 180^o - \dfrac{\widehat{D} + \widehat{C}}{2}$
$\Leftrightarrow \widehat{AQB} + \widehat{DPC} = 360^o - \dfrac{\widehat{A} + \widehat{B} + \widehat{D} + \widehat{C}}{2} = 360^o - \dfrac{360^o}{2} = 180^o$
mà $\widehat{AQB} + \widehat{DPC} = 360^o - (\widehat{QNP} + \widehat{QMP})$
⇒ $\widehat{QNP} + \widehat{QMP} = 360^o - 180^o = 180^o$