WLOG Assume that $BD=1$ for convenience Letting $AB=k$, using the quadratic formula it is easy to see that $k^2=k+(\frac{1}{2})^2$, or $k^2-k-\frac{1}{4}=0$ $\rightarrow k=\frac{1+\sqrt{1+1}}{2}=\frac{1+\sqrt{2}}{2}$. Now, using basic Pythagoras it is easy to see that $BC=\frac{\sqrt{2}}{2}$ Construction: Extend $CD$ past $D$ a distance of $0.5$, and do the same for line $BC$ past B. Draw the perpendiculars of both of these lines and mark where the meet. We will now show that this point is on $AE$. Let $Q,R$ be the points such that $QB=0.5$ and $RD=0.5$ and the points $C,B,Q$ and $C,D,R$ being collinear in that order, and $S$ is where the perpendiculars of $BQ, DR$ meet. Since $QCRS$ is a square, it is obvious that $QS=\frac{1+\sqrt{2}}{2}$, and as $QS$ and $AB$ intersect at $45$ degrees, for $S$ to be on $AE$ we require by Pythagoras that $AB=\sqrt{(\frac{1}{2})^2+(\frac{1}{2})^2}+\frac{1}{2}$, which is obviously true. Similarly with $RS$, and we also have that $AS=SE$. Now, we can cut the pentagon into 3 shapes by cutting it along $SQ$ and $SR$ to remove the 2 triangles containing points $A$ and $B$, and we can place them such that the right angle of $A$ overlaps that of point $Q$ and similarly with point $B$ on $R$. We then have a square with side length of $\frac{1+\sqrt{2}}{2}$ with vertices $QCRS$, which is our desired result. The attachment is a little out of proportion, apologies in advance

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