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 $E \neq O$, 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 $|EX| = |EY|$. $\square$