a) Point $B$ lies on the radical axis, $BD$, of circles $c_1$ and $c_2$, therefore $BK \cdot BA = BL \cdot BC$, which indicates that the quadrilateral $AKLC$ is cyclic. So is $ACMN$. It follows that $\angle LKB = \angle ACB$. The angle between the tangent at $A$ to the circle $c_1$ (which is also tangent to $C$) with line $AB$ subtends arcs $AK$ of circle $c_1$ and $AB$ of circle $C$, and therefore $\angle ANK = \angle ADB$. We obtain that $NK \parallel BD$ and, similarly, $ML \parallel BD$. (This fact also follows from the homotheties that transform circles $c_1$, and $c_2$, respectively, into $C$.) Finally, $\angle NKL = 180^\circ - \angle AKN - \angle LKB = 180^\circ - \angle ABD - \angle LKB = 180^\circ - \angle ACD - \angle ACB = \angle BAD$. Similarly, $\angle MNK = \angle BAD$, which leads to the conclusion. Alternatively, one could have noticed that line segments $ML$, $PQ$ and $[NK]$ share the same perpendicular bisector.

b) The radical axes of circles $c_1$, $c_2$, $C$ (one for each pair of circles), i.e., the tangent line to $C$ at $A$, the tangent line to $C$ at $C$ and the line $BD$ are not all parallel, therefore they are concurrent in the radical center. Diagonal $BD$ is then a symmedian of triangle $ABC$, which means that quadrilateral $ABCD$ is harmonic. (One can also use this fact to give a different proof to a).)

As $NPQK$ is an isosceles trapezoid, it follows that $\angle NAP = \angle QAK$, which means that rays $(AP$ and $(AQ$ are isogonal with respect to angle $\angle DAB$). But $ABCD$ being harmonic, $AP$ is a symmedian of triangle $DAB$, therefore $AQ$, which is its isogonal, is the median.

*Alternative Solution.* One could use a property of harmonic quadrilaterals that was put into evidence by the Danube Mathematical Competition from the same year: If $R$ is the midpoint of diagonal $BD$ of a harmonic quadrilateral $ABCD$, then $\angle ARD = \angle CRD$.

One proves that $R$ is the only point on the line segment $BD$ that has the property from above. Next, one proves that point $Q$ does have the property, which makes it the midpoint of $BD$.