Goutham wrote: The diagonals of a convex quadrilateral $ABCD$ are perpendicular to each other and intersect at the point $O$. The sum of the inradii of triangles $AOB$ and $COD$ is equal to the sum of the inradii of triangles $BOC$ and $DOA$. $(a)$ Prove that $ABCD$ has an incircle. $(b)$ Prove that $ABCD$ is symmetric about one of its diagonals. Let $s_{XYZ}, r_{XYZ}$ denote the semi-perimeter and inradius respectively, of an arbitrary $\triangle XYZ$. Now, it is well known that if $\triangle XYZ$ is right-angled at $X$, then $r_{XYZ} = s_{XYZ} - YZ$. $(a)$ We use the above fact repeatedly. \begin{align*} AB+CD &= (s_{AOB} - r_{AOB}) + (s_{COD} - r_{COD}) \\ &= s_{AOB} + s_{COD} - (r_{AOB} + r_{COD}) \\ &= s_{AOB} + s_{COD} - (r_{COB} + r_{AOD}) = s_{AOB} + s_{COD} - ( s_{COB} - BC + s_{AOD} - AD) \\ &= BC + AD + (s_{AOB} + s_{COD} - s_{COB} - s_{AOD}) \\ &= \frac 32 (BC + AD) - \frac 12 (AB+CD)\end{align*} which immediately gives the result by Converse of Pitot's theorem. $(b)$ We have, $AB^2 + CD^2 = AD^2+BC^2$ since the quadrilateral has perpendicular diagonals. Now, combining with $AB+CD = AD+ BC$, we easily get $AB \cdot CD = AD \cdot BC$ leading to the set of lengths $ { AD,BC }$ being the same as the set ${AB,CD }$. Now, this directly gives us that $AD = AB$ and $BC = CD$, or $AD = CD$ and $BC = AB$. In both cases, the quadrilateral is symmetric about one of its diagonals. We are done.