Clearly the coordinates of all points are linear when $D$ moves (linearly) along the line $AC$. So it suffices to prove the claim for two distinct cases. But in the cases $D=A$ and $D=C$ the claim is trivially true. Done.

Another way: If we rotate the whole figure by $90^\circ$ around $A$, $B \mapsto C, D \mapsto E$ and hence $P \mapsto R$. So $AP=AR$ and $\angle RAP=90^\circ$. Moreover, since $R$ is the midpoint of $CE$ and $\angle EAC=\angle EQC=90^\circ$, $EQAC$ is cyclic with circumcenter $R$. So $QR=AR=AP$. But similarly $ABQD$ is cyclic with circumcenter $P$ and so $PQ=AP=AR=QR$ so $ARQP$ is a square.

Finally: (Explicit) Coordinates. W.l.o.g. take $A=(0,0), B=(0,1), C=(1,0)$. Then $D=(x,0)$ for some $x \in [0,1], E=(0,-x), R=\left(\frac{1}{2}, -\frac{x}{2}\right), P=\left(\frac{x}{2}, \frac{1}{2}\right), Q=\left(\frac{x+1}{2}, \frac{1-x}{2}\right)$. Then one can easily check that $ARQP$ is indeed a square.

the problem also appeared 16 years ago, in May Olympiad (Olimpiada de Mayo) 2002 L2 as P3

$Lemma:$ If $ABC$ is a right triangle, right in $B$, and $D$ is the intersection of the median of $B$ to $AC$, so $AD \cong DC \cong BD$. $Proof: AD \cong DC$ by definition. Tracing a perpendicular from $D$ to $BC$, intersecting at $E$. It's not hard to notice that $DE$ is a midsegment of $ABC$. So the triangle $BED \cong CDE$ by side-angle-side congruence, which give us $BD \cong DC$ as we desired. The triangle $ABC$ of the question is necessarily a 90-45-45 triangle. Applying the $Lemma$ in the triangle $ABD$, give us that $BP \cong PD \cong PA.$ Note that $AD \cong AE \Leftrightarrow \angle ADE = 45 \Leftrightarrow \angle QDC = 45 \Leftrightarrow \angle DQC = 90 \Leftrightarrow \angle BQD = 90$. Applying the $Lemma$ in the triangle $AQD$, we get that $PA \cong BP \cong PD \cong PQ$, so $PA \cong PQ$. Applying the $Lemma$ in the triangles $ACE$ and $EQC$, we get $AR \cong ER \cong RC \cong RQ$. Since $AD \cong AE, BA \cong AC$, and $\angle BAC = \angle EAC$, we have that $AEC \cong ADB$. So $PA \cong PQ \cong PD \cong \frac{BD}{2} \cong \frac{EC}{2} \cong RE \cong RA \cong RQ $ We concluded that $AP \cong PQ \cong QR \cong RA$ Let $\angle ABD = \beta$ and $\angle BDA = \alpha$ $\angle AER = \angle EAR = \alpha \Rightarrow \angle DAR = 90- \alpha = \beta.$ $\angle PAD = \angle PDA = \alpha.$ So the $\angle PAR = \angle RAD + \angle PAD = \alpha + \beta = 90$ Look at the $\angle PBQ = PQB = 45 - \angle PAB = 45-\beta,$ once that $PB \cong PQ$. And so $\angle DPQ = 2(45-\beta) = 90 -2\beta$. The $ \angle APD = \angle BAP + \angle PBA = 2 \beta $. So $ \angle APQ = \angle APD + \angle DPQ = 2\beta +90 - 2\beta = 90$. $AP \cong PQ \cong QR \cong RA$ and $\angle PAR = \angle APQ \Leftrightarrow APQR$ is a square.

This problem is very similar to ISL G1 2020