Orthohedron Problem: Finding BE When CG = 1/2 HG

To solve this problem, we'll proceed with the following steps:

• Identify the necessary orthohedron relationship provided: $CG = \frac{1}{2} HG$.
• Apply the Pythagorean theorem to determine the necessary dimensions and lengths.
• Use the given conditions to find the expression for $BE$.

Now, let's work through each step:

1. **Identify and understand the relations**: The problem states $CG = \frac{1}{2} HG$.

2. **Dimension considerations**: Let $x$ represent the unknown variable related to the dimension of the orthohedron. Relating it with known components, use relations like heights, bases, and the values directly available from vertex connections.

• Let $E = (0, 0, 0)$ be the origin, as vertex positioning might suggest traditional base positioning.
• Let the dimension lengths be specified on pertinent axes $(x, y, z)$, giving appropriate value correlations for verifying $CG = \frac{1}{2} HG$.

3. **Apply Pythagorean theorem**: Use the modification in dimension $CG$, compared to the general diagonal calculation within the 3D rectangular prism to derive the length.

This leads us to identifications for size:

• Determine the relationship for triangle spans in 3D.
• In terms of required placement, this involves appropriate usage of lengths squared.
• Geometrically identify direct solutions for sum and placement to achieve: $cg = x \frac{\sqrt{5}}{2} + \frac{5\sqrt{5}}{2}$.

4. **Calculation**: Use calculations based on the simplifications offered by Pythagoras and algebraic manipulations laid out from vertex placements and prism features.

**Conclusion**: Therefore, the length $BE$ is $x\frac{\sqrt{5}}{2}+\frac{5\sqrt{5}}{2}$.

The answer is: $BE = x \frac{\sqrt{5}}{2} + \frac{5\sqrt{5}}{2}$.