We first show that $AVMS$ is cyclic. Since the triangle $ABC$ is isosceles, $AM \perp BC$, and since $NP$ is midline, $V$ is the midpoint of $AR$. Then $MV$ is the $M$-median of the right triangle $AMR$, so $VM = VA = VR$. Hence $V$ lies on the perpendicular bisector of $AM$. By hypothesis, $V$ also lies on the internal bisector of $\angle ASM$, so, in the circle $AMS$, $V$ is the midpoint of the arc $AM$ not containing $S$. Consequently, $AVMS$ is cyclic, as stated. Similarly, $AUMT$ is cyclic.

We now prove that $H$ lies on the circle $NUS$. Since $AVMS$ is cyclic, $\angle ASV = \angle MAR$. By hypothesis, $\angle MAR = \angle BAQ$, so $AB \parallel SV$. Similarly, $AC \parallel TU$. Then $\angle SVP = \angle APV = 90^\circ - \frac{1}{2}\angle BAC$, and $\angle AUH = 90^\circ - \angle UAV = 90^\circ - \frac{1}{2}\angle BAC$, so $\angle SVP = \angle AUH$.

From the triangles $SUV$ and $AUN$,
$$\frac{SU}{SV} = \frac{\sin(90^\circ - \frac{1}{2}\angle BAC)}{\sin \angle SUV} = \frac{AU}{AN}.$$
Since $\angle UAM = \angle UAH = \angle VAN$ and $\angle AUH = 90^\circ - \frac{1}{2}\angle BAC = \angle ANV$, the triangles $AUH$ and $ANV$ are similar.
Hence $AN/AU = NV/UH$, so $SU/SV = UH/VN$. Since $\angle SVN = 180^\circ - \angle SVP = 180^\circ - \angle AUM = \angle SUH$, the triangles $SUH$ and $SVN$ are similar, so $\angle USH = \angle VSN$, whence $\angle USV = \angle HSN$.
Finally, $\angle USV = \angle MAV = 90^\circ - \angle AVU = \angle HUV$, so $\angle HSN = \angle HUV = \angle HUN$. Consequently, $H$ lies on the circle $NUS$, as stated. Similarly, $H$ lies on the circle $PVT$. This ends the proof.