Question 149
We are given a triangle *ABC* such that $\angle BAC \lt 90^\circ$. The point $D$ is on the opposite side of the line $AB$ to $C$ such that $|AD| = |BD|$ and $\angle ADB = 90^\circ$. Similarly, the point $E$ is on the opposite side of $AC$ to $B$ such that $|AE| = |CE|$ and $\angle AEC = 90^\circ$. The point $X$ is such that $ADXE$ is a parallelogram. Prove that $|BX| = |CX|$.