Given AF = 10, AD = DB, CF = FE = EB. Determine the length of GF.

Since $AD = DB$, $D$ is the midpoint of $AB$. Thus, $\frac{AD}{DB} = 1$.
Since $CF = FE = EB$, let $CF = x$. Then $FE = x$ and $EB = x$.
The length of $CB = CF + FE + EB = x + x + x = 3x$.
So, $\frac{CB}{CF} = \frac{3x}{x} = 3$.

Consider $\triangle ABF$ and the transversal line $DGC$. The line $DGC$ intersects side $AB$ at $D$, side $AF$ at $G$, and the extension of side $BF$ at $C$.
According to Menelaus' Theorem:

Substitute the values from Step 1 into the equation:

This implies $GA = 3 \cdot FG$.
We are given that $AF = 10 \text{ cm}$.
Also, $AF = AG + GF$.
Substitute $AG = 3 \cdot FG$ into the equation for $AF$:

The length of $GF$ is $\boxed{\text{2.5 cm}}$.

Since $ABCD$ is a parallelogram, its opposite sides are equal and parallel.

• $AB = CD$ (Opposite sides of parallelogram $ABCD$).
• $AB \parallel CD$ (Opposite sides of parallelogram $ABCD$). Since $AB \parallel CD$ and $AC$ is a transversal, the alternate interior angles are equal:
• $\angle BAE = \angle DCF$. We are given that $BE \perp AC$ and $DF \perp AC$.
• $\angle BEA = 90^\circ$ and $\angle DFC = 90^\circ$.

In $\triangle ABE$ and $\triangle CDF$:

• $\angle BEA = \angle DFC = 90^\circ$ (Angle)
• $\angle BAE = \angle DCF$ (Angle, alternate interior angles)
• $AB = CD$ (Side, opposite sides of parallelogram) By the Angle-Angle-Side (AAS) congruence criterion, $\triangle ABE \equiv \triangle CDF$.

Since $\triangle ABE \equiv \triangle CDF$, their corresponding parts are equal (CPCTC).

• $BE = DF$.

We are given that $BE \perp AC$ and $DF \perp AC$.
If two lines are perpendicular to the same line, then they are parallel to each other.
Therefore, $BE \parallel DF$.

In quadrilateral $BEDF$, we have shown that one pair of opposite sides ($BE$ and $DF$) are both equal in length ($BE = DF$) and parallel ($BE \parallel DF$).
A quadrilateral with one pair of opposite sides that are both parallel and equal in length is a parallelogram.
Therefore, $\boxed{\text{BEDF is a parallelogram}}$.