IRL_ABooklet_2023

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 $|BX|=|CX|$.



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Alternative solution.

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 $$\begin{array}{rl}\angle BDX& =90^{\circ }-\angle ADX=90^{\circ }-(180^{\circ }-\angle DAE)=\angle DAE-90^{\circ }\\ & =\angle DAE-(45^{\circ }+45^{\circ })=\angle DAE-\angle DAB-\angle EAC\\ & =\angle BAC.\end{array}$$ 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 $|BX|=|CX|$.