Inscribed circles in Quadrilaterals

Question

In a convex quadrilateral $ABCD$, we have points $P \in AB; R\in AD$. Lines $PD$ and $RB$ intersect at a point $S$. Assume we can inscribe a circle in each of the quadrilaterals $APSR$ and $CDSB$. How do I prove that we can inscribe a circle in $ABCD$

Answer 1

Hint: Simply mark the four points of tangeny of the circle inscribed in $APSR$ and the other four points of tangeny of the circle inscribed in $ CDSB$. Then assign edge-length variables to the segments formed by adding these four points to the rest of the points given to you (you get a lot of segments). Have in mind that the tangents to the same circle from a given point are equal! Also, notice that that the two interior tangents to the two incircles are equal (the ones through the point $S$)! Then carefully form all the equations (sums of lengths) for these edge-length variables and after simplifying them you will obtain at the end that $AB + CD = BC + CD$ which is a necessary and sufficient condition for the existence of an incircle in $ABCD$.