Posted before under exactly the same title concurrent and parallel.

Quote: Points $P$ and $Q$ are taken sides $AB$ and $AC$ of a triangle $ABC$ respectively such that $\hat{APC}=\hat{AQB}=45^{0}$. The line through $P$ perpendicular to $AB$ intersects $BQ$ at $S$, and the line through $Q$ perpendicular to $AC$ intersects $CP$ at $R$. Let $D$ be the foot of the altitude of triangle $ABC$ from $A$. Prove that $SR\parallel BC$ and $PS,AD,QR$ are concurrent. Solution: WLOG, assume, $A-B-P$ and $A-C-Q$. Let $SP \cap QR=Y$ and $PC \cap BQ=X$ $\implies$ $APYQ$ is cyclic, $\angle SPR$ $=$ $135^{\circ}$ $=$ $\angle SQR$ $\implies$ $SPQR$ is cyclic. $\angle BPC$ $=$ $\angle BQC$ $\implies$ $BPQC$ is cyclic $$\angle QSR=\angle QPR=\angle QBR \implies BC||SR$$Let $SR \cap AY=Z$ $$\angle YAQ=90^{\circ}-\angle AYQ=90^{\circ}-\angle APQ=45^{\circ}-\angle CBQ=45^{\circ}-\angle RSQ=\angle ZRQ $$Hence, $AZ \perp SR$ $\implies$ $D \in AZ$ $\implies$ $AD, SP, QR$ concurrent at $Y$

We have $$\angle SPR=90^\circ+45^\circ=180^\circ-45^\circ=180^\circ-\angle SQR\implies S,P,R,Q\text{ are concyclic}.$$So, $$\angle QBC=\angle QPC=\angle QPR=\angle QSR\implies \boxed{SR\|BC}.$$Let $AD\cap SR=F$. $$\begin{cases}\angle AFS=\angle APS=90^\circ\implies \text{$A,F,P,S$ are concyclic} \\ \angle AFR=\angle AQR=90^\circ\implies \text{$A,F,Q,R$ are concyclic.}\end{cases}$$So, by radical center theorem $QR,AF,SP$ concur. $\blacksquare$