Let $M$ be the center of side $AC$ triangle $ABC$. Points $P$ and $Q$ are placed on sides $AM$ and $MC$ ($P-AM$,$Q-MC$) so that $PQ=\frac{AC}{2}$.... Let $M$ be the midpoint of the side $AC$ of the triangle $ABC$. Points $P$ and $Q$ are placed on the segments $AM$ and $MC$ respectively so that $PQ=\dfrac{AC}{2}$. The circumcircle of the triangle $ABQ$ cuts $BC$ at $X$ and the circumcircle of the triangle $BCP$ cuts $AB$ at $Y$. Prove that $BXMY$ is a cyclic quadrilateral.
I think that we need to use the cyclic quadrilateral $BXQA$ in some way and maybe prove a congruence using $PQ$ as $PQ=\dfrac{AC}{2}=AM=CM$. I've tried this and more, but to no avail. If you see anything, please feel free to share your idea.

 A: $\bf{\text{A Useful Corollary}}:$
The corollary of the converse of the theorem Euclid III. 22 states that the exterior angle formed by producing a side of a cyclic quadrilateral is equal to its interior opposite angle.

For brevity, let $\angle CAB = \alpha$, $\angle ABC = \omega$, $\angle BCA = \theta$, and $\angle YMP = \phi$. We also add two segments, i.e. $PY$ and $QX$, to the sketch provided by OP.
Since $MA = PQ$, we have $MA-MP = PQ-MP$, which means.
$$PA = QM. \tag{1a}$$
Arguing similarly, we can show that
$$CQ = MP. \tag{1b}$$
Let us do some angle chasing with the view to prove the triangles $ABC$, $AYP$ and $CQX$ are equiangular.
Consider the cyclic quadrilaterals $ABXQ$ and $BCPY$, which are inscribed in the circumcircles of $\triangle ABQ$ and $\triangle BCP$ respectively. We are applying the abovementioned corollary to these two cyclic quadrilaterals
According to this corollary, the exterior angle $QXC$ formed by producing the side $BX$ of the cyclic quadrilateral $ABXQ$ is equal to its interior opposite angle $QAB$, i.e.
$$\measuredangle QXC = \alpha. \tag{2a}$$
Application of the same corollary shows that the exterior angle $AYP$ formed by producing the side $BY$ of the cyclic quadrilateral$ BCPY$ is equal to its interior opposite angle $BCP$, i.e.
$$\measuredangle AYP = \theta. \tag{2b}$$
It follows from (2a) and (2b) that each of the three triangles $ABC$, $AYP$ and $CQX$ has two angles equal to $\alpha$ and $\theta$. Therefore, the three triangles are equiangular. Since the remaining angle $ABC$ of $\triangle ABC$ is equal to $\omega$, the remaining angles $CQX$ of $\triangle CQX$ and $YPA$ of $\triangle AYP$ are also equal to $\omega$. Hence, we shall write,
$$\measuredangle XQM = \measuredangle MPY = 180^o - \omega. \tag{3}$$
Since the triangles $AYP$ and $CQX$ are equiangular, according to the theorem Euclid VI. 4, their corresponding sides are proportional. Therefore, we have,
$$\dfrac{QX}{PA}=\dfrac{CQ}{YP}. \tag{4}$$
When we substitute the values of $PA$ and $CQ$ from (1a) and (1b) in (4), we get,
$$\dfrac{QX}{QM}=\dfrac{MP}{YP}\quad\Longrightarrow\quad \dfrac{QX}{MP}=\dfrac{QM}{YP}. \tag{5}$$
Now, consider the two triangles $XQM$ and $MPY$. Using (3)and (5), and applying the theorem Euclid VI. 6, we can show that the triangles $XQM$ and $MPY$ are equiangular. Therefore, we have,
$$\measuredangle MXQ = \measuredangle YMP = \phi. \tag{6}$$
As shown below, we can determined the angle $QMX$ of $\triangle XQM$ using (3) and (6).
$$\measuredangle QMX = 180^o - \measuredangle XQM - \measuredangle MXQ = 180^o -\left( 180^o - \omega \right) - \phi = \omega -\phi \tag{7}$$
Finally, using (7), we can find the angle $XMY$, which is the opposite angle of $ABC$ of the quadrilateral $BXMY$.
$$\measuredangle XMY = 180^o - \measuredangle YMP - \measuredangle QMX = 180^o - \phi - \left(\omega -\phi \right) = 180^o - \omega.$$
This means the the two opposite angles $\measuredangle XMY$ and $\measuredangle XABC$ of the quadrilateral $BXMY$ are supplementary. Therefore, according to the converse of the theorem Euclid III. 22, the quadrilateral $BXMY$ is cyclic.
