Let $ABCD$ be a parallelogram. $H$ is the Orthocenter of $\triangle ABC$. Points $P\in BC $ and $Q\in AD$ such that $PQ \parallel AB$ with $(PQ)$ passes through $H$, and points $R\in AB $ and $S\in CD$ such that $RS \parallel BC$ with $(RS)$ passes through H. Show that $PRQS$ is cyclic.
Here is a diagram
If you look carefully, you can see that $\angle QRP$ and $\angle QSP$ are about $90$, so maybe we can try to prove that thing.
$$\angle QRP=180-\angle ARQ-\angle BRP=180-\angle RQH-\angle RPH$$
Same thing with $$\angle QSP=180-\angle HPS-\angle HQS$$
Hence $$\angle QRP+\angle QSP=360-\angle QRH-\angle RPH-\angle HPS-\angle HQS$$ But to get that $QRP+QSP+180$ You have to prove $\angle RPS+ \angle RQS=180$. It's the same thing if you tried to prove $\angle SRP=\angle SQP$. Maybe this problem is beyond Angle Chasing?
Actually, you could prove this by proving $QH\times HP=RH\times HS$ without looking at the angles.
This is equivalent to $AR/RH=PC/PH$ since $AR=QH$ and $SH=PC$.
Now, notice that $H$ is orthocenter, we have $\angle BAH=90^\circ-\angle ABC=\angle BCH$, so we have $\angle PAH=\angle PCH$. Furthermore, since $AR\parallel PH$ and $RH\parallel PC$ we have $\angle ARH=\angle CPH$. Thus, we have $\triangle ARH\sim\triangle CPH$, and therefore $AR/RH=PC/PH$