(treated during the session)

Let $ABCD$ be a quadrilateral, $P$ the intersection of $(AC)$ and $(BD)$. Suppose that $\widehat{CAD}=50^{\circ}$, $\widehat{BAC}=70^{\circ}$, $\widehat{DCA}=40^{\circ}$, and $\widehat{ACB}=20^{\circ}$. Calculate the angle $\widehat{CPD}$.
We notice that $\widehat{B A D}+\widehat{D C B}=\widehat{B A C}+\widehat{C A D}+\widehat{D C A}+\widehat{A C B}=70^{\circ}+50^{\circ}+40^{\circ}+20^{\circ}=180^{\circ}$.

Therefore, $A, B, C, D$ are concyclic. Thus $\widehat{P D C}=\widehat{B D C}=\widehat{B A C}=70^{\circ}$. So $\widehat{C P D}=180^{\circ}-\widehat{P D C}-$ $\widehat{D C P}=180^{\circ}-70^{\circ}-40^{\circ}=70^{\circ}$.

Therefore, $\widehat{C P D}=70^{\circ}$.