AC + CE = -1/4 + 1/2 sqrt[(257/16) - 4(a^2 + b^2)] + 1/2 sqrt[(257/16) - 4(c^2 + d^2)].

Vo Duc Dien wrote: AC + CE = -1/4 + 1/2 sqrt[(257/16) - 4(a^2 + b^2)] + 1/2 sqrt[(257/16) - 4(c^2 + d^2)]. No, it is really not. $AC+CE = \frac{ad(\sqrt{64a^2 -1} + \sqrt{64d^2 -1}) +d(\sqrt{4-a^2} + \sqrt{4-d^2})}{8ad}$ which is not so nice. I don't know the answer. But I observed some good points about the construction. The distance of $B$ to $AC$ is $\frac18$. The distance of $D$ to $CE$ is $\frac18$. To construct such a pentagon, $AC/2 < 1-1/8=7/8$ and $CE/2 < 1-1/8=7/8$. $AC<7/4$ and $CE<7/4$. $28^\circ< \cos^{-1}(7/8)< \angle CAE< \sin^{-1}(7/8) < 62^\circ$. If $a$ is given, it is possible to find $b,c,d$ in terms of $a$. $\frac{7+\sqrt 15}4 < AC+CE < 2\sqrt 2$ and according to Geometer's Sketchpad $AC+CE \approx a+b+c+d - \tfrac{1}{10}$.

Let $AC=x$ and $CE=y$. We have $BE^2 = 4-a^2$, $AD^2 = 4 - d^2$, and $x^2 + y^2 = 4$. Ptolemy in $ABCE$: $(ay+2b)^2 = x^2(4-a^2)$ $a^2y^2 + 4b^2 + 4aby = 4x^2 - a^2x^2$ Ptolemy in $ACDE$: $(2c+dx)^2 = y^2(4-d^2)$ $d^2x^2 + 4c^2 + 4cdx = 4y^2 - d^2y^2$ Adding these two side by side $4b^2 + 4c^2 + a^2y^2 + a^2x^2 + d^2x^2 + d^2y^2 + 4aby + 4cdx = 4x^2 + 4y^2$ substitute $x^2 + y^2 = 4$, $ab=cd=1/4$ $4a^2 + 4b^2 + 4c^2 + 4d^2 + x + y = 16$ $x+y = 16 - 4a^2 - 4b^2 - 4c^2 - 4d^2 = 4(4-a^2-b^2-c^2-d^2)$.$\blacksquare$

From the law of Cos and $ab=cd=\frac{1}{4}$, $|AC|^2=a^2+b^2-\frac{1}{2}cosB$, $|CE|^2=c^2+d^2-\frac{1}{2}cosD$, Then, $4=a^2+b^2+c^2+d^2-\frac{1}{2}(cosB+cosD)$ From the triangle $ACE$, $-cosB=\frac{|CE|}{2}$ and $-cosD=\frac{|AC|}{2}$,so we get: $4=a^2+b^2+c^2+d^2+\frac{1}{4}(|AC|+|CE|)$, $|AC|+|CE|=16-4(a^2+b^2+c^2+d^2)=4[4-(a^2+b^2+c^2+d^2)]$