smc

Let $BE=x$. Let $JA=y$. Because $\angle FKH=\angle EJK=\angle AGJ=\angle DHG$ and $\angle FHK=\angle EKJ=\angle AJG=\angle DGH$, $△KEJ,△JAG,△GDH,△HFK$ are all similar. Using proportions and the pythagorean theorem, we find $$EK=xy$$ $$FK=\sqrt{1-y^2}$$ $$EJ=x\sqrt{1-y^2}$$ Because we know that $BE+EJ+AJ=EK+FK=1$, we can set up a systems of equations and solving for $x$, we get $$x+x\sqrt{1-y^2}+y=1⟹x=\frac{1-y}{1+\sqrt{1-y^2}}$$ $$xy+\sqrt{1-y^2}=1⟹x=\frac{1-\sqrt{1-y^2}}{y}$$ Now solving for $y$, we get $$\frac{1-y}{1+\sqrt{1-y^2}}=\frac{1-\sqrt{1-y^2}}{y}⟹y(1-y)=(1-\sqrt{1-y^2})(1+\sqrt{1-y^2})⟹y-y^2=y^2⟹y=\frac{1}{2}$$ Plugging into the second equations with $x$, we get $$x=2\left(1-\sqrt{1-\frac{1}{4}}\right)=2\left(\frac{2-\sqrt{3}}{2}\right)=\boxed{2-\sqrt{3}}$$