I read the following Pitot's theorem:
A quadrilateral ABCD is tangential if and only if $AB+CD=AD+BC$, where $AB$ means the length of side $AB$.
How can I prove it. I mean, the case $ABCD$ is tangential $\implies AB+CD=AD+BC$ is easy because there is a theorem that if A lies outside the circle and B,C are points where the tangent of a given circle passes through A then $AB=AC$. But if $AB+CD=AD+BC$, why do we can draw a quadrilateral around the circle?
Without loss of generality, we may suppose that $AD$ is the minimum side.
(1) When $AB=AD$, we have $BC=CD$. In this case, letting $O$ be the intersection point of $AC$ and the bisector of $\angle B$, there exists a circle $O$ which is inscribed in the $ABCD$.
(2) When $AB\gt AD$, we have $BC\gt CD$.
Here, take $E$ on the side $AB$, $F$ on the side $BC$ such that $$AE=AD, \ \ \ CF=CD.$$ Since we have $$AD+BC=AB+CD\iff AB-AD=BC-CD,$$ we have $$BE=BF.$$ Hence, we can see that each bisector of $\angle A,\angle B,\angle C$ is the perpendicular bisector of the side $DE,EF,FD$ of $\triangle DEF$ respectively.
Hence, since the bisectors of $\angle A,\angle B,\angle C$ intersect at one point, the quadrilateral $ABCD$ is circumscribed to a circle.
