Let quadrilateral $ABCD$ be inscribed in $(O)$. Let $I_1, I_2, I_3, I_4$ be the center of the circle inscribed in triangles $ABD, ADC, DBC, ABC$ respectively. $I_1I_3$ cuts $AC$ at$ P$, $I_2I_4$ cuts $BD$ at $Q.$ $I_1I_3 $cuts $I_2I_4$ at $S$.
- Prove that $OS\perp PQ$.
- Prove that $I_1I_2I_3I_4$ is a rectangle.
We have $\widehat{I_2CI_3} = \frac{\widehat{BCD}-\widehat{ACD}}{2} = \frac{\widehat{BCA}}{2}$
$\widehat{I_2DI_3} = \frac{\widehat{ADB}}{2} = \widehat{I_2CI_3} $
So we have $I_2I_3DC$ as inscribed quadrilateral. Similarly, we also have $I_1I_4BA,I_4I_3CB,I_1I_2DA$ as inscribed quadrilaterals.
$\Rightarrow \widehat{I_4I_3I_2} = 360^\circ-\widehat{I_2I_3C} - \widehat{I_4I_3C} = 90^\circ$
Similarly, we have $I_1I_2I_3I_4$ is a rectangle.
That's all I can infer, hope to get help from everyone. Thanks very much !