Potential inversion geometry problem: Prove that quadrilateral $DPMQ$ is cyclic Let $AD$ be the angle bisector of acute triangle $ABC$, and let $M$ be the midpoint of $AD$. Let $P$ be the point on $BM$ such that $\angle APC = 90^{\circ}$, and let $Q$ be the point on $CM$ such that $\angle AQB = 90^{\circ}$. Prove that $DPMQ$ is a cyclic quadrilateral.
My thoughts: It looks like I can use the inversion of circle $M$ with radius $MD$, Then I can try to show $B$ goes to $P$ and $C$ goes to $Q$. But I am not able to finish it.

 A: Sketch. Consider point $F$ on the line $AC$ such that $BF\parallel AD$. Then, prove that $CM$ passes through the midpoint of $BF$. After that, check that circumcircle of $AQB$ passes through the midpoint of $BF$ as well. Conclude that $\angle AQM=\angle MAC$ and similarly $\angle APM=\angle BAM$. It follows that $MA^2=MD^2=MQ\cdot MC=MP\cdot MB$ and, hence, $DPMQ$ is a cyclic quadrilateral.
A: HINT.- This way calculations are somewhat tedious but they are straightforward.
$(1)$ Let a triangle $\triangle{ABC}$ of sides $a,b,c$ and see the following points in figure $1$ below
$$B(0,0),C(a,0),A(a_1,a_2),D\left(\frac{ac}{b+c},\frac{ab}{b+c}\right),M\left(\frac{ac+a_1(b+c)}{2(b+c)},\frac{ab+a_2(b+c)}{2(b+c)}\right), X\left(\frac{a_1}{2},\frac{a_2}{2}\right),Y\left(\frac{a_1+a}{2},\frac{a_2}{2}\right)$$ where $X,Y$ are the midpoints of sides $c$ and $b$ respectively and $D$ is given by the theorem of the angle bisector.
$(2)$ Now the circle centered at $Y$ with radius $\dfrac b2$ gives the point $P$ as intersection with line $BM$ and the  circle centered at $X$ with radius $\dfrac c2$ gives the point $Q$ as intersection with line $MC$ (see figure $2$. In other words $P$ is given by the system $$\begin{cases}(2x-a_1-a)^2+(2y-a_2)^2=b^2\\\dfrac yx=\dfrac{ab+a_2(b+c)}{ac+a_1(b+c}\end{cases}$$ and $Q$ by $$(2x-a_1)^2+(2y-a_2)^2=c^2\\ \begin{cases}\dfrac yx=\dfrac{-(ab+a_2(b+c))}{2ab+ac-a_1(b+c)}\end{cases} $$
$(3)$ Having the four points $M,D,P,Q$, write the equation of the circle passing by three of them and verify that the other belongs to this circle. For a direct way you can use the known formula
$$ \begin{vmatrix}
x_1^2+y_1^2&x_1&y_1&1\\
x_2^2+y_2^2&x_2&y_2&1\\
x_3^2+y_3^2&x_3&y_3&1\\
x_4^2+y_4^2&x_4&y_4&1
\end{vmatrix}=0$$ where the $(x_i,y_i); i=1,2,3,4$ are the coordinates of the four points $M,D,P,Q$ whatever be the order.

