Invalid URL First, we prove it for a cyclic quadrilateral. Since $\angle AI_{A}B = 90+\frac{1}{2}\angle ADB = 90+\frac{1}{2}\angle ACB = \angle AI_{B}B$ and so on, we get that the quadrilaterals $AI_{A}I_{B}B, BI_{B}I_{C}C, CI_{C}I_{D}D, DI_{D}I_{A}A$ are all cyclic. Then we can calculate $\angle I_{D}I_{A}I_{B}= 360-\angle AI_{A}I_{B}-\angle AI_{A}I_{D}=$ $\angle ABI_{B}+\angle ADI_{D}=$ $\frac{1}{2}\angle ADC+\frac{1}{2}\angle ABC = 90$. We conclude that $I_{A}I_{B}I_{C}I_{D}$ is a rectangle. Therefore, for each point P in the plane, $PI_{A}^{2}-PI_{D}^{2}= PI_{B}^{2}-PI_{C}^{2}$: you just have to draw the projection of P on a side and apply Pithagora's theorem. Take P=O and the case "cyclic quadrilateral" is done. Remark: from Euler's formula, $OI^{2}= R^{2}-2rR$, we get the very interesting fact that the sum of the inradii of $ABC,ADC$ equals the sum of the inradii of $BCD,BAD$. Now proving the general case is just a combinatorial problem. In fact it'easy to see that you can transform any triangulation to an other by doing this operation: - take two adiacent triangles in the triangulation that form the cyclic quadrilateral ABCD, and swap its diagonal. We have proved that this operation does not change the sum of the distances from O to the incenters. We prove that for any triangulations and any vertex X of the polygon, using that operation you can obtain the triangulation "all vertex linked to X". This can be done by induction: of no vertex is linked to X, then the vertices Y,Z adjacent to X are linked, take the quadtilateral XYZT where YZT forms another triangle of the triangulation, and change the diagonal from YZ to ZT. Once you have done this, the diagonal XT divides the polygon in two polygons, and continue doing this on each of the two polygons, you will end up with the desired configuration. Nice problem

as for quadrilateral,it's a question appeared in China TST.