AustriaMO2011

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.