Help with proving the following quadrilateral is cyclic Let AB be the diameter of a semi circumference. A point M is marked over the semi circumference, and a point K over AB. Draw circumference with center P that passes by the points A, M and K, and another circumference with center Q that passes by the points M, K and B. Show that M, K, P and Q are concyclic.
What I´ve tried so far is, as the sides of the quadrilateral are radii, the triangles APK and BQK are isosceles. Then, the line that connects the centers of both circles is also the height of both triangles that passes trough the side MK. That shows that the diagonal of the quadrilateral intersect at right angles, which turns the problem to showing that the median of one of the triangles intersects at a right angle the side oposite of the other triangle. 
Also, the problem reduces to showing that the triangle QMP has a right angle at M. This can be shown by proving tha te angle MPQ is equal to MKQ.
 A: There is no need to draw so many circles. Let's keep it simple. $AB$ being the diameter of a semicircle containing $M$ is equivalent to $\angle AMB=90$. 
Now, the circumcenter $P$ is the intersection of the bisectors of $AK$ and $MK$. And the circumcenter $Q$ is the intersection of the bisectors of $MK$ and $KB$. Let's denote $D,E,F$ the midpoints of $AK$, $KB$, $MK$ respectively. So, we have $\angle PDK=\angle PFK=\angle QFK=\angle QEK=90$. 
So, far we just translated all of the information that we had. Now, here is the solution:
1) Since $FD,FE$ are midsegments, we have $FD\mathbin{/\mkern-6mu/} MA$ and $FE\mathbin{/\mkern-6mu/}MB$ . Since $MA\perp MB$, we conclude that $FD\perp FE$ as well, i.e. $\angle DFE=90$
2) Since $\angle DFE=90$, we have that $\angle PFD+\angle EFQ=90$
3) Note that $PFKD$ and $QFKE$ are cyclic quadrilaterals. Hence, $\angle DKP=\angle PFD$ and $\angle EKQ=\angle EFQ$. In particular, $\angle DKP+\angle EKQ=90$.
4) Since $\angle DKP+\angle EKQ=90$, we conclude that $\angle PKQ=90$. 
5) Since $PQ$ is the bisector of $MK$, we have by symmetry that $\angle PMQ=\angle PKQ=90$. In particular, $MPKQ$ is a cyclic quadrilateral.
A: This might have a simple solution computing angles, but with so many circles and demanding to prove that points were concyclic, this problem looks ideal for inversion.
If you invert with center at $K$ and radius $1$ (the radius is not important), let $A'$, $B'$ and $M'$ be the inverses of $A$, $B$ and $C$, respectively. 
Call $C$ the first circle, $C_1$ the circle through $AMK$ and $C_2$ the circle through $BMK$.
The inverse of $C_1$ is the line $M'A'$, and the inverse of $C_2$ is the line $M'B'$.
Since $C$ is symmetric with respect to $AB$, then $C$ transforms into a circle $C'$ that is also symmetric with respect to the line $A'B'$.
Therefore the lines $M'A'$ and $M'B'$ are perpendicular, because $A'B'$ is a diameter of $C'$.
Now, the image $P'$ of $P$ is the reflection $K_1$ of $K$ with respect to the line $M'B'$. The image $Q'$ of $Q$ is the reflection of $K$  by the line $M'A'$.

Explanation of the claim above. Note that $P$ and $\infty$ are inverses of each other with respect to $C_1$. Therefore $P'$ and the inverse of $\infty$ by the inversion with center at $K$ are symmetric with respect to the inverse of $C_1$, which is the line $M'A'$. But the inverse of $\infty$ is the center of inversion, and that is $K$. For that reason $P'$ is the reflection of $K$ with respect to the line $M'A'$. Likewise goes the argument of why $Q'$ is the reflection of $K$ with respect to the line $M'B'$.

Since these two lines are perpendicular, then $P'$, $M'$, and $Q'$ are collinear.
Therefore, their pre-images are concyclic with $K$.

Alternative ending of the inversion.
Since the lines $M'A'$ and $M'B'$ are perpendicular (since $A'B'$ is a diameter of $C'$). Then the circles $C_1$ and $C_2$ intersect perpendicularly. Therefore $PM$ is perpendicular to $QM$. By symmetry of $PMQK$ it follows that $PK$ is perpendicular to $QK$. Therefore, $PMQK$ is cyclic.
