Solution:

Let us call $r$ and $s$ respectively the external bisectors of $D\widehat{A}C$ and $D\widehat{B}C$. Let us construct the points $C'$ and $D'$ respectively as the reflection of $C$ with respect to $s$ and as the reflection of $D$ with respect to $r$. Since $r$ is the external bisector, we have that $C', B$ and $D$ are collinear and moreover by construction $C'B = CB$; hence $C'D = C'B + BD = BC + BD$. In the same way $D'C = AD + AC$. Let us now call $\beta = B\widehat{P}C$ which is equal by construction to $B\widehat{P}C'$, and in the same way $\alpha = A\widehat{P}D = A\widehat{P}D'$. Finally let us call $\gamma = C\widehat{P}D$.

Let us now consider the triangles $C'PD$ and $CPD'$. We have by construction $PD' = PD$ and $PC' = PC$; therefore the two triangles are equal if and only if the two angles at $P$ are equal or, equivalently, if and only if the third side is equal. But the first condition says that $C'\widehat{P}D = 2\beta + \gamma = 2\alpha + \gamma = C\widehat{P}D'$, which is equivalent to $\alpha = \beta$, while the second condition says that $CD' = C'D$ which, by what was shown above, is equivalent to $AD + AC = BC + BD$.