IMO Team Selection Test 2

We consider the configuration as in the figure, where $E$ is at least as close to $B$ as it is to $C$. The proof in the case of the configuration in which this is the other way around, is analogous.

Let $O$ be the centre of $\Gamma$. The angle between the line $BC$ and the common tangent in $B$ is on the one hand, by the inscribed angle theorem (tangent case), equal to $\angle BED$, and on the other hand equal to $\angle BAC$. So $\angle BED=\angle BAC$. Analogously, we show that $\angle CED=\angle BAC$, so $\angle BEC=\angle BED+\angle CED=2\angle BAC=\angle BOC$, where we use the inscribed angle theorem to derive the last step. Therefore $E$ lies on the circle that passes through $B$, $O$, and $C$.

If $E=O$, then we're done, as $|EX|$ and $|EY|$ then both are the radius of the circle.

So suppose that $E\ne O$, then in the configuration considered, $BEOC$ is a cyclic quadrilateral. Then $\angle BEO=180^{\circ }-\angle BCO$. In the isosceles triangle $BOC$, we have $\angle BCO=90^{\circ }-\frac{1}{2}\angle BOC=90^{\circ }-\angle BAC$, so $\angle BEO=180^{\circ }-(90^{\circ }-\angle BAC)=90^{\circ }+\angle BAC$. Hence $\angle DEO=\angle BEO-\angle BED=90^{\circ }+\angle BAC-\angle BAC=90^{\circ }$. Therefore $EO$ is perpendicular to $DE$ and therefore also perpendicular to chord $XY$, from which follows that $E$ is the midpoint of $XY$. We conclude that $|EX|=|EY|$. $\square$