It is well known that the angle bisector in $C$ and the bisector of the side $AB$ intersect in a point on the circumcircle of a triangle $ABC$. Let this point be $M$. It is certainly equidistant from $A$ and $B$. We now consider the triangle $AIM$. Naming the angles in $A$, $B$ and $C$, $\alpha$, $\beta$ and $\gamma$ as usual, we note that $\angle MAI = \frac{\alpha}{2} + \angle BAM = \frac{\alpha}{2} + \angle BCM = \frac{\alpha}{2} + \frac{\gamma}{2}$. Furthermore, $\angle IMA = \angle CMA = \angle CBA = \beta$, and we therefore have $\angle MIA = 180^\circ - \angle IMA - \angle AIM = 180^\circ - \beta - (\frac{\alpha}{2} + \frac{\gamma}{2}) = \frac{\alpha}{2} + \frac{\gamma}{2}$. We see that the triangle $AIM$ is isosceles, and we have $MA = MI$. $M$ is therefore the mid-point of $k$.

If we name the mid-points of $AP$ and $BQ$ $U$ and $V$ respectively, we see that triangles $MUC$ and $MVC$ are certainly congruent, since they both have angles of $90^\circ$ and $\frac{\gamma}{2}$ and the common hypotenuse $CM$. We therefore have $MU = MV$. Since $MA = MP = MB = MQ$, we now see that the triangles $AMP$ and $BMQ$ are both isosceles with the same side lengths and the same altitudes, and are therefore also congruent, from which it follows that $AP = BQ$ holds.

The quadrilateral $AQBP$ is therefore inscribed and has two sides of the same length, and is therefore a trapezoid, as claimed.

We also see that neither $AC$ nor $BC$ can be a tangent of $k$. If either were a tangent, the two (congruent) triangles $AMP$ and $BMC$ would both degenerate to segments perpendicular to $AC$ and $BC$ respectively. Both lines would therefore be tangent to $k$, and since the tangent segments would therefore be of equal length, it would follow that $ABC$ is isosceles, which is a contradiction to the assumption that it is not. This completes our proof.