Diagonals of the cyclic quadrilateral $ABCD$ are intersecting each other at the point $S$. Orthogonal projections of $S$ on the edges of the quadrilateral lie on those edges respectively. Prove those projections are vertices of a tangent quadrilateral.
My attempt: $$\measuredangle CAB=\measuredangle CDB\;\land\;\measuredangle BSA=\measuredangle DSC\;\implies\;\Delta ABS{\sim}\Delta CDS$$ Analogously:$$\Delta DAS{\sim}\Delta BCS$$ Let $E,F,G,H$ be the orthogonal projections on the edges $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$ of $ABCD$ respectively.
The given right-triangles: $$\Delta HAS,\Delta SAE, \Delta BSE,\ldots$$
I wanted to use the mentioned above to show: $$|EF|+|GH|=|FG|+|HE|,$$ but it went unsuccessfully.
So far, I've learned:
The center of the incircle of a tangent quadrilateral is the intersection point of the four angle-bisectors.
While drawing the polygons and circles in GeoGebra, I saw the diagonals of $ABCD$ and the four angle-bisectors of $EFGH$ intersect each other at $S$, so $S$ is the center of the incircle of $EFGH$.
However, I haven't managed to prove it.
Picture (Where I also drew a right-triangle $\Delta RQU$ because I thought I could use:$\measuredangle QSR=2\measuredangle QUS$):
May I ask for advice on solving this task? Thank you in advance!
Because $$\measuredangle LFS=\measuredangle LBS=\measuredangle ABD=\measuredangle ACD=\measuredangle SFG,$$ which says that $FS$ is a bisector of $\angle LFG.$
Can you end it now?