For a parallelogram $ABCK$, $AD$ and $CE$ are the altitudes of $\triangle ABC$ $\triangle AHQ\sim\triangle DHP$, $~$ $\triangle SHC\sim\triangle RHE$ $RH=BP=\frac{EH.BC}{CE}$ , $~$ $HS=PC=\frac{CH.BC}{CE}$ Likewise, $PH=\frac{HD.AB}{AD}$ , $~$ $HQ=\frac{AH.AB}{AD}$ $RH.HS=\frac{EH.CH.BC^{2}}{CE^{2}}$ , $~$ $PH.HQ=\frac{HD.AH.AB^{2}}{AD^{2}}$ $~$ and $~$ $~$ $\frac{AD}{AB}=\frac{CE}{BC}$ Thus, $~$ $RH.HS=PH.HQ$ $~$ and then $~$ $\square PRQS$ is cyclic.

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@Headhunter: Nice solution! I like similar triangles, for it always gives us a neat solution. Anyway, I found another approach, let's see if it's valid. First, it's clear from the condition of the problem that H must be inside the $\triangle ABC$. And since H is the orthocenter of the $\triangle ABC$, it's known that the undirected-angle $\angle AHC = 180^{\circ} - \angle ABC = 180^{\circ} - \angle ADC$, which implies that the quadrilateral $AHCD$ is cyclic. This means $\angle SDH = \angle CAH =\angle HBP$. Now we know that $\triangle DHS \sim \triangle BHP$. Then, by the ratios, $\frac{DS}{SH} = \frac{BP}{PH}$. This then implies $QH\cdot HP = RH \cdot HS$ as the above solution has done.

My solution: Since $PG \slash \slash AB$ and $RS \slash \slash BC$, we have : $\angle ARS = \angle QHS = \angle HPC$ We also have $ \angle PHC =\angle BEC= \frac{\pi}{2} = \angle ADB = \angle ARH$ $\Rightarrow$ $\triangle ARH \sim \triangle PHC$ $\Rightarrow$ $ \frac{AR}{RH}= \frac{PC}{PH}$. $ (1)$ $AQRH$ and $HPCS$ are parallelograms. $\Rightarrow$ $AR = HQ$ and $ HS = PC$. $(2)$ From $(1)$ and $(2)$, we get: $RH \cdot HS = PH \cdot HQ$ By the reciprocal of the power of a point, we can conclude that P, Q, R and S are one a circle.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%204.pdf p. 18-20. Sincerely Jean-Louis

It suffices to show that $AP \cdot PD = AR \cdot RB$ by power of a point from $H$. Let $\overline{BE} \perp \overline{AD}$ and $\overline{DF} \perp \overline{AB}$ with $E, F$ lying on $\overline{AD}, \overline{AB}$ respectively. Then $$AR \cdot RB = \frac 1{\sin A} \cdot HB \cdot HE = \frac 1{\sin A} \cdot HD \cdot HF = AP \cdot PD,$$so we are done.

This was in 106 geometry problems. I couldn’t draw the picture lol.