Proof. Note that $\overline{AQ} \parallel \overline{BP}$, as both are perpendicular to $\overline{CD}$. Since lines $AP$ and $BQ$ are distinct, lines $AQ$ and $BP$ are distinct.
By symmetric reasoning, we get that $AQCPBR$ is a hexagon with opposite sides parallel and concurrent diagonals as $\overline{AP}$, $\overline{BQ}$, $\overline{CR}$ meet at $T$. This implies that the hexagon is centrally symmetric about $T$; indeed
$$\frac{AT}{TP} = \frac{TQ}{BT} = \frac{CT}{TR} = \frac{TP}{AT}$$
so all the ratios are equal to $+1$.
Next, $\overline{PD} \perp \overline{BC} \parallel \overline{QR}$, so by symmetry we get $D$ is the orthocenter of $\triangle PQR$. This means that $T$ is the midpoint of $\overline{DH}$ as well. □

Corollary
The configuration is now symmetric: we have four points $A$, $B$, $C$, $D$, and their reflections in $T$ are four orthocenters $P$, $Q$, $R$, $H$.
Let $S$ be the centroid of $\{A, B, C, D\}$, and let $O$ be the reflection of $T$ in $S$. We are ready to conclude:
Claim — $A$, $B$, $C$, $D$ are equidistant from $O$.
Proof. Let $A'$, $O'$, $S'$, $T'$, $D'$ be the projections of $A$, $O$, $S$, $T$, $D$ onto line $BC$.
Then $T'$ is the midpoint of $\overline{A'D'}$, so $S' = \frac{1}{4}(A' + D' + B + C)$ gives that $O'$ is the midpoint of $\overline{BC}$.
Thus $OB = OC$ and we're done. □