Constructing Escher's Template for the Tetrahedron 5-Compound 
Consider the above figure. The lengths of all the outer line segments are known, and are specified below. We know that angle $A$ is 60 degrees and that line segment $|CD|$ extends $|AD|$.
This information, as far as I can figure, is not enough to actually construct this figure, though it should be enough to derive all required information.
I also have a suspicion that $|BC| = |CD| = |BD|$, which if true would be enough information, but thus far I've been unable to prove it.
The more specific facts can be found at: http://mathworld.wolfram.com/Tetrahedron5-Compound.html
But all we really need to know is this:
The side lengths are:
$|AB| = |AE| = 1$
$|DE| = \phi^{-1} \approx 0.618033988749895$
$|BC| = |CD| = \phi^{-1}/\sqrt2 \approx 0.437016024448821$
and what I've already mentioned, where $\phi$ denotes the golden ratio.
It's all about the line segment $|CD|$; that is the real unknown. I basically just need one single piece of information, other than what I've already got, and then I won't have a problem solving the problem. It doesn't really matter what that information is either, except perhaps angle D. I'm not quite sure that would help.
So that's it, this being my first post, I do hope I've not messed up too badly. Also I'm certain this is not a hard problem, but I've been struggling with it on a off for a couple of years now, believe it or now, so some help would be much appreciated.
Best regards.
 A: Let's first check whether you have enough information. You need 5 points in the plane, each with 2 real degrees of freedom. But you don't care about position or rotation of the figure as a whole, which accounts for 3 real degrees of freedom. Which means you need 7 measurements to hope for full specification. You have 5 lengths, 1 angle and 1 collinearity. Could work.
You can pick coordinates and use them to express your distance and collinearity constraints.
\begin{align*}
A&=(0,0) & B&=(1,0) & C&=(C_x,C_y) & D&=(D_x,D_y) & E&=(\tfrac12,\tfrac12\sqrt3)
\end{align*}
\begin{align*}
(C_x-1)^2+C_y^2 &= \frac1{2\phi^2} = \left(\frac{\phi^{-1}}{\sqrt2}\right)^2\\
(C_x-D_x)^2+(C_y-D_y)^2 &= \frac1{2\phi^2} \\
(D_x-\tfrac12)^2+(D_y-\tfrac12\sqrt3)^2 &= \frac1{\phi^2} \\
\frac{C_x}{C_y} &= \frac{D_x}{D_y}
\end{align*}
This can be rewritten as a set of polynomial equations. Then typical elimination strategies like Gröbner bases or Resultants can be used to find solutions. But I prefer to leave these things to a computer algebra system, which will tell me that there exist two possible solutions:
\begin{alignat*}{3}
C_x &= \tfrac14\sqrt5 + \tfrac12 &&\approx 1.0590169943749474 &
C_x &\approx 0.9369017680762171 \\
C_y &= \tfrac14\sqrt3 &&\approx 0.4330127018922193 &
C_y &\approx 0.4324368378770998 \\
D_x &= \tfrac18\sqrt5 + \tfrac38 &&\approx 0.6545084971874737 &
D_x &\approx 0.5401122930754872 \\
D_y &= \tfrac14\sqrt{-\tfrac32\sqrt5 + \tfrac92}
&&\approx 0.2676165673298174\qquad &
D_y &\approx 0.2492944939101796
\end{alignat*}
Your picture shows $C$ slightly right of $B$, so presumably the solution with $C_x>1$ is the one you want. Which is good, because the other solutions are roots of a polynomial of degree $10$ and more, with no radical expression available. This also demonstrates that a ruler and compass solution for that second part is impossible. Since going from the problem statement both solutions would be permissible, reaching either using compass and straightedge is going to be tricky.
You can use these coordinates to confirm that yes, indeed $\lvert BC\rvert=\lvert CD\rvert=\lvert BD\rvert$. So another approach would have been to assume this as the additional piece of information, then compute positions for everything else from that and check that the resulting coordinates satisfy all requirements. More specifically, draw $A,B,E$ then construct $D$ as intersection of two circles cooresponding to the distances to $B$ and $E$. Then draw the triangle $BCD$ and check that line $CD$ passes through $A$.
