$\angle AQC=\angle APC=90^o$ As $B,P,Q,C$ lies on circle and $BP=CQ$ then $BQ \parallel CP$ so $BQ \perp AP$. But $AP \perp QS$ so $S$ is point of intersection of $BQ$ and $AP$ $BH \perp AQ \to BH \parallel QC$. $\angle QDC= \angle PAB, \angle QCD = \angle HBA$ so $\triangle DQC = \triangle AXB$ So $AX=DQ$ and $XB=QC=BP$. But $BS$ is height for isosceles $\triangle XBP$ so $S$ is midpoint $XP \to XS=PS$

RagvaloD wrote: $\angle AQC=\angle APC=90^o$ As $B,P,Q,C$ lies on circle and $BP=CQ$ then $BQ \parallel CP$ so $BQ \perp AP$. But $AP \perp QS$ so $S$ is point of intersection of $BQ$ and $AP$ Then $X$ is the orthocentre of $ABQ$, so both results follow from the Reflections of the Orthocentre Lemma ($P$ is on $(ABQ)$ and on the altitude from $A$; and the antipode of $B$ on $(ABQ)$ is $D$).

Since $AQ$ is the diameter, $\angle{ASQ}=90^{\circ}$. Since $BP=CQ, \angle{PAB}=\angle{CAQ}, \angle{BAC}=\angle{SAQ}$. Since $\angle{ABC}=\angle{ASQ}=90^{\circ}, \angle{AQS}=\angle{ACB}$. Which implies $B,S,Q$ lie on the same line. $\angle{SAQ}=90-\angle{SQA}=90-\angle{ACB}=90-\angle{BPX}, BP=BX$. Since $BQ\bot XP, XS=PS$ Second one notice $AB=CD, BX=CQ, \angle{BAX}=\angle{CDQ}$. $SSA$ is a legit way to determine congruency for obtuse triangle, so $\triangle{BAX}\cong \triangle{CDA}$ which implies the result

First we prove that $AX$=$DQ$ , denote the intersection of perpendicular from $B$ to $AQ$ with $AQ$ As $T$ so $\angle TXA$ = $90 - $ $\angle TAX$ = $\frown BP/2 + \frown DQ/2$ = $\angle XAB$ + $\angle XBA$ = $\frown BP/2$ + $\angle XBA$ so $\angle XBA$ = $\angle QCD$ , $\angle XAB$ = $\angle QDC$ , $DC$ = $AB$ hence $\triangle XAB$ = $\triangle QDC$ so we get $AX$ = $DQ$ Since $\angle DAP$ = $\angle QPA$ , $DQ$ and $AP$ are parallel so $ADQP$ is a Isosceles trapezoid so $AD$ = $PQ$ and since $ADQX$ is a parallelogram so $AD$ = $QX$ so $QX$ = $QP$ so we get that $QS$ is the perpendicular bisector of $PX$ and therefore $XS$ = $PS$.

Claim 1: BQS collinear QBC = QAC = PAB So AP line and BQ line make equal angles in same sense with lines AB amd BC, so they are perpendicular (i) it suffices to show BQ is perpendicular bisector of XP so it suffices to show BP = BX equivalent to CQ = BX. CQA = BGA = 90 (G is BX cap AQ) so BX parallel to CQ. So it suffices to show BXQC parallelogram. BX parallel CQ, AB parallel CD and AB=CD. So ABX = BCQ. We also have CQ=BP so CDQ = BAP = BAX. So BXA congruent CQD. So AX parallel DQ. Now consider a translation of length AD along AD. A does to D, B goes to C and to BXA goes to CQD (As congruent triangles in same sense and 2 points are mapped so 3rd must be). So XQ=AD. And XQ parallel to AD and hence BC. So BCQX and ADQX are parallelogram which shows this part and the next.