What's the measure of the $\angle PQS$?

Question

Fr referene:In an inscribed quadrilateral $ABCD, ~P,~ Q, ~and~ S$ are the incenters of the triangles $ABD,~ ABC, ~and ~BCD$, respectively. Calculate $ \measuredangle PQS $.

My progress:

Here's the drawing I made and the relationships I found..

Answer 1

Japanese Theorem (source:https://www.cut-the-knot.org/Curriculum/Geometry/CyclicQuadrilateral.shtml#explanation)

From $ ΔBCD, ∠BI_{BCD}C = 90° + \frac{∠BDC}{2}$

Similarly, from: $ΔABC\rightarrow ∠BI_{ABC}C = 90° + \frac{∠BAC}{2}$ $ABCD$ is cyclic: $∠BDC = ∠BAC.$

$\therefore ∠BI_{BCD}C = ∠BI_{ABC}C.$

$BI_{ABC}I_{BCD}C \rightarrow$ is cyclic.

$(1) ∠BCI_{BCD} + ∠BI_{ABC}I_{BCD} = 180^\circ$

Similarly the quadrilateral $AI_{ABD}I_{ABC}B$ is cyclic.

$(2) ∠BAI_{ABD} + ∠BI_{ABC}I_{ABD} = 180^\circ$.

From(1) and (2) $\implies ∠BI_{ABC}I_{BCD} + ∠BI_{ABC}I_{ABD} = 360^\circ \frac{∠BCD}{2} - \frac{∠BAD}{2} = 270^o\implies$

$(3) ∠I_{ABD}I_{ABC}I_{BCD} = 90°$