Force equation for pulley FBD with symmetric tension

[STEMucator]

Homework Statement

I'm having a bit of trouble understanding something from this:

http://gyazo.com/88a466b57d8516d2df1b12008947be43

Homework Equations

The Attempt at a Solution

I understand how to get the components of reaction at $A$ as well as the tension in segment $BE$.

What I'm having trouble with is the FBD of the pulley used to determine the tension in $DEC$. I'm not quite sure how they have come up with a part of the force equation, namely:

$\sum F_z = 0 \Rightarrow 2(\frac{4}{\sqrt{96}})T - \frac{1}{\sqrt{5}}(1677.05) = 0$

The term $2(\frac{4}{\sqrt{96}})T$ is what is confusing. I understand you need twice the tension since it's symmetric, but how on Earth are they getting $\frac{4}{\sqrt{96}}$?

I figure they are applying $\frac{T_z}{T} = \frac{4}{\sqrt{96}}$ to a triangle, but I'm not seeing the triangle.

 

[SteamKing]

If you look at the diagram, write the coordinates of points C, D, and B.

When the pulleys are under tension, one can see that CDE make a triangle and a plane. You can also assume that the line BE will also lie in that same plane. You can determine the position of the point E by drawing a projection of the plane CDE in the y-z coordinate plane, and extending it to the line AB on the y-axis. Once the coordinates of point E are known, you can calculate the length of line DE, which is Sqrt (96) feet.