Let $T$ be the reflection of $A$ with respect to $X$. By the converse of the Power of a Point Theorem, it follows that the quadrilateral $ABTC$ is cyclic. If the line parallel to $BC$ through $T$ intersects the circumcircle of $ABC$ again at $U$, and lines $AU$ and $BC$ meet at $Y$, then $Y \in [BC]$ and $BY \cdot CY = AY \cdot UY = AY^2$, thus, in order for the point $X$ to be unique, it is necessary that $T = U$, i.e. $T$ is the midpoint of the arc $BC$, in other words $X$ is the foot of the angle bisector from $A$.

From here one can finish the proof in several ways. One can use the formula for the length of the angle bisector in a triangle, or Stewart's Theorem.

If $AX = l$, $BX = x$, $CX = y$, then $b^2x + c^2y = l^2a + axy$. It is known that $x = \frac{ac}{b+c}$, $y = \frac{ab}{b+c}$. It follows that
$$l^2 = xy \Leftrightarrow b^2x + c^2y = 2axy \Leftrightarrow (b+c)^2 = 2a^2 \Leftrightarrow b+c = a\sqrt{2}.$$