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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 $△PQR$. 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 $A^{\prime }$, $O^{\prime }$, $S^{\prime }$, $T^{\prime }$, $D^{\prime }$ be the projections of $A$, $O$, $S$, $T$, $D$ onto line $BC$. Then $T^{\prime }$ is the midpoint of $\overline{A^{\prime }{D}^{\prime }}$, so $S^{\prime }=\frac{1}{4}(A^{\prime }+D^{\prime }+B+C)$ gives that $O^{\prime }$ is the midpoint of $\overline{BC}$. Thus $OB=OC$ and we're done. □