Let $ \widetilde{A}, $ $ \widetilde{D} $ be the isogonal conjugate of $ A, $ $ D $ WRT $ \triangle BCD, $ $ \triangle ABC, $ respectively. Clearly, $ \widetilde{A} $ and $ \widetilde{D} $ are symmetry WRT $ BC, $ so one of the intersection $ U $ of $ \odot (PQR) $ and $ \odot (XYZ) $ lies on $ BC. $ Let $ V $ be the second intersection of these two circles, then note that $ ( A,B,P,Q ) $ and $ ( B,D,Y,Z ) $ are concyclic we get $$ \measuredangle ZVQ = \measuredangle ZYU + \measuredangle UPQ = \measuredangle ZDB + \measuredangle BAQ = 2\measuredangle ABD \ , $$hence $ V $ lies on the 9-point circle of $ \triangle ABD. $ Let $ M $ be the midpoint of $ BH. $ Since $ M, $ $ Q, $ $ V, $ $ Z $ are concyclic (the 9-point circle of $ \triangle ABD $), so we conclude that $$ \measuredangle MVQ = \measuredangle BAQ = \measuredangle UPQ = \measuredangle UVQ \Longrightarrow M \ \text{lies on} \ UV \ . $$

EDIT: 4321st post woo

I think that this solution is a bit different from others posted here. Not that hard a problem with the knowledge of circumrectangular hyperbolas, especially with geogebra by your side mofumofu wrote: Let $ABCD$ be a non-cyclic convex quadrilateral. The feet of perpendiculars from $A$ to $BC,BD,CD$ are $P,Q,R$ respectively, where $P,Q$ lie on segments $BC,BD$ and $R$ lies on $CD$ extended. The feet of perpendiculars from $D$ to $AC,BC,AB$ are $X,Y,Z$ respectively, where $X,Y$ lie on segments $AC,BC$ and $Z$ lies on $BA$ extended. Let the orthocenter of $\triangle ABD$ be $H$. Prove that the common chord of circumcircles of $\triangle PQR$ and $\triangle XYZ$ bisects $BH$. First we clear up our notation a bit. Consider the equivalent problem: China TST 2017 #2, morally correct notation wrote: Let $ABCP$ be a non-cyclic quadrilateral and $\mathcal{H}$ denote the rectangular hyperbola circumscribing them. Let $H_B$ be the orthocenter of triangle $ABP$. Let $\omega_1$ be the pedal circle of $P$ wrt $\triangle ABC$ and $\omega_2$ be the pedal circle of $A$ wrt $\triangle BCP$. Prove that the radical axis (or common chord) of $\omega_1, \omega_2$ passes through the mid-point of $\overline{BH_B}$. For convenience, suppose $P$ lies in the interior of $\triangle ABC$. The general case can be accounted for by directing angles accordingly. Now let $Z$ be the center of $\mathcal{H}$ and $P_A, P_B, P_C$ be the projections of $P$ on $\overline{BC}, \overline{CA}, \overline{AB}$ respectively. Let $\odot(P_AP_BP_C)$ meet the sides $\overline{BC}, \overline{CA}, \overline{AB}$ again at $Q_A,Q_B,Q_C$ respectively. Let $M$ be the midpoint of $\overline{BH_B}$ and $P_B'=\overline{AP} \cap \overline{BH_B}$. Then $H_B \in \mathcal{H} \implies Z \in \odot(P_BMP_B')$. Claim. $\overline{ZQ_A}$ is the common chord of $\omega_1, \omega_2$ (in fact these two points lie on both the circles). (Proof) Observe that $Z$ is a common point due to the fundamental theorem and $Q_A \in \omega_1$. So let $A'$ be the isogonal conjugate of $A$ wrt $\triangle BPC$; and $P'$ be the isogonal conjugate of $P$ wrt $\triangle ABC$. Then $A', P'$ are symmetric in $\overline{BC}$. Consequently, $Q_A$ lies on the pedal circle $\omega_2$ of $A'$ wrt $\triangle BPC$, as desired. $\blacksquare$ Finally, we see $$\angle P_BZQ_A=\angle P_BP_AB=90^{\circ}-\angle ABP$$and $$\angle P_BZM=\angle P_BP_B'B=90^{\circ}-\angle ABP$$hence $M$ lies on $\overline{ZQ_A}$, as desired. $\blacksquare$ P.S. 1400th post!

ZR-BC=U then XZU=XAR=XYC so UXYZ cyclic, similarly QXU collinear so RQU=RZX=RAC=RPC so UPQR cyclic hence ZVQ=ZVU-UVQ=ZYU-UPQ=180-ZDB-QB=ZMQ so MVQ+QVU=180-MZQ+QAB=180 so q.e.d

Notice that $Z,Q,X,R$ all lies on the circle with diameter $AD$. Denote this circle by $\omega$.Let $N$ be the midpoint of $AD$. Let $M$ be the midpoint of $BH$. Suppose $ZX$ and $QR$ meet at $G$, and $ZR$ and $QX$ meet at $I$. CLAIM. Both $I$ and $G$ lies on the common chord Proof. Applying radical axis theorem to $(PQR)$, $(XYZ)$ and $\omega$ we see that $G$ is the radical center of the three circles hence lying on the common chord. Now $$\angle QIR=\angle ZRQ-\angle XQR=\angle BAQ-(90^{\circ}-\angle ACD)=\angle QPC-\angle RPC=\angle QPR$$hence $I$ lies on $(PQR)$. By symmetry it lies on $(XYZ)$ as well. This justfies our claim. Now notice that $\angle MQB+\angle NQD=\angle MBQ+\angle NQD=90^{\circ}$. Therefore $MZ$ and $MQ$ are both tangent to $\omega$. Let $ZQ$ and $DX$ meet at $J$. Then $J$ lies on $ZQ$, the polar of $M$ w.r.t. $\omega$. Hence by La Hire's theorem, $M$ lies on the polar of $J$. w.r.t. $\omega$, that is, $GI$ by Brokard's theorem. This shows that the common chord of the two circles bisect $BH$.

Pascal's on $\odot (AZRDXQ)$, $ZR \cap QX \cap PY=I'_2$. Let $U=AD \cap BC$, $V=ZQ \cap RX$, and assume $AB<CD$. By simple angle chasing, we see $$\angle RI_2B=\angle RAD-\angle ZDA+\angle DUC=\angle ADC-\angle DAB+\angle DAB+\angle CBA-\pi=\angle ADC+\angle CBA-\pi=\pi-\angle AQD+\pi-\angle PQA-\pi=\pi-\angle AQD-\angle PQA=\pi-\angle PQR$$ so $\odot (I'_2PQR)$ is cyclic, and similarly, $\odot (I'_2YXZ)$ is cyclic. Let $M$ be the midpoint of $BH$. Then, $MH=MZ=MQ=MB$, and $$\angle MZB= \angle MBZ=\frac{\pi}{2}-\angle ZAD=\angle ZDA$$ so $MZ, MQ$ are tangent to circle $\odot (AZRDXQ)$, and $M$ lies on the polar of $V$ w.r.t. $\odot (AZRDXQ)$, which, by Brokard's theorem, is $I_1I_2$, the radical axis of $\odot (XYZ)$ and $\odot (PQR)$.

Let $K$ be the orthocenter of $\triangle ACD$, and let $M$ and $N$ be the midpoints of $BH$ and $CK$. Claim: $\overline{HK}, \overline{BC}$ and $\overline{ZR}$ are concurrent one of the intersections of $(PQR)$ and $(XYZ)$, say $S$. Proof. Let $U=\overline{KR}\cap\overline{HZ}$, and $V=\overline{CR}\cap\overline{BZ}$. Then as $D$ is the orthocenter of $\triangle AUV$, $UV\parallel CK\parallel BH$. Hence $\triangle KRC$ and $\triangle HZB$ are perspective and the concurrency follows. Now it is left to show that this point lies on both circles. Since $ZAQR$ and $BAQP$ are cyclic, by Miquel's theorem it follows that $S$ lies on $(PQR)$. Similarly, $S$ lies on $(XYZ)$ so we are done. $\square$ Now as $BH\parallel CK$ it follows by homothety that $\overline{MN}$ passes through $S$. Consider a rectangular circumhyperbola $\mathcal{H}$ passing through $A,B,C,D,H$. Obviously, this passes through $K$ as well. Let $T$ be the center of this hyperbola. Then by the fundamental theorem, this must be the other intersection point of $(PQR)$ and $(XYZ)$ (Configuration issues can be dealt without much difficulty). Since $BH$ and $CK$ are parallel chords of a conic, it follows that $\overline{MN}$ passes through $T$, so it must be the radical axis of $(PQR)$ and $(XYZ)$ as desired. $\square$

casey theorem for distance $H,B$ with radical axis of circles $\odot XYZ$,$\odot PQR$.

Denote by $\Omega$ the circle passing through $A,Z,R,D,X,Q$. By Pascal's theorem on $(ZRDQXA)$, we have that $L:=ZR \cap QX$ lies on line $BC$. Simple angle chasing indicates that $L$ is one of the intersections of circles $(PQR)$ and $(XYZ)$. Consider the radical axis of $\Omega$, $(PQR)$ and $(XYZ)$, we have $S:=XZ\cap RQ$ lies on the radical axis of $(PQR)$ and $(XYZ)$, which we denote by $l$. Thus now we conclude that $l$ is line $SL$, so we only have to prove that the $M$ , the midpoint of segment $BH$, lies on $SL$. To finish, it is well-known that $M$ is the pole of line $QZ$ with respect to $\Omega$, therefore apply Pascal's theorem on $(ZZRQQX)$, and then we are done!

$\text{a good problem for training the \textit{Pascal theorem}, but it is a bit easy for China TST}$ geogebra solution link