Since $ADXE$ is a parallelogram, we have $\angle ADX = \angle AEX$. This implies that $\angle XDB = 90^\circ - \angle ADX = 90^\circ - \angle AEX = \angle CEX$. Since triangle $ADB$ is isosceles and $ADXE$ is a parallelogram, we have $|DB| = |DA| = |XE|$. Similarly, $|EC| = |EA| = |XD|$. We conclude that triangles $XDB$ and $CEX$ are congruent by SAS and so $|BX| = |CX|$.

First we note that $\angle DBA = \angle DAB = \angle ECA = \angle EAC = 45^\circ$. Since $ADXE$ is a parallelogram, we have $\angle ADX + \angle DAE = 180^\circ$. Thus
$$\begin{align*} \angle BDX &= 90^\circ - \angle ADX = 90^\circ - (180^\circ - \angle DAE) = \angle DAE - 90^\circ \\ &= \angle DAE - (45^\circ + 45^\circ) = \angle DAE - \angle DAB - \angle EAC \\ &= \angle BAC. \end{align*}$$
Next, note that $\frac{|AB|}{|BD|} = \sqrt{2} = \frac{|AC|}{|AE|} = \frac{|AC|}{|DX|}$. Thus $\frac{|BD|}{|DX|} = \frac{|AB|}{|AC|}$.
Combining $\angle BDX = \angle BAC$ and $\frac{|BD|}{|DX|} = \frac{|AB|}{|AC|}$, we deduce that triangles $DBX$ and $ABC$ are similar. It follows that $\angle DBX = \angle ABC$ and so $\angle XBC = \angle DBA = 45^\circ$. A similar argument shows that $\angle XCB = 45^\circ$. Thus the triangle $XBC$ is isosceles, and $|BX| = |CX|$.