Write down AH : HG. Calculate t if the area of the parallelogram is a maximum. (NOTE: Area of a parallelogram = base x ⊥ height)

Since $\text{DECF}$ is a parallelogram, $\text{DE} \parallel \text{FC}$. As $\text{F}$ lies on $\text{BC}$, this implies $\text{DE} \parallel \text{BC}$.
Because $\text{DE} \parallel \text{BC}$, $\triangle \text{ADE}$ is similar to $\triangle \text{ABC}$ ($\triangle \text{ADE} \sim \triangle \text{ABC}$).

From the parallelogram, $\text{DE} = \text{FC}$.
We are given $\text{BF} : \text{FC} = 2 : 3$.
This means $\text{FC} = \frac{3}{2+3} \text{BC} = \frac{3}{5} \text{BC}$.
Therefore, $\text{DE} = \frac{3}{5} \text{BC}$.

For similar triangles $\triangle \text{ADE}$ and $\triangle \text{ABC}$, the ratio of their corresponding heights is equal to the ratio of their corresponding bases:

Substitute $\text{DE} = \frac{3}{5} \text{BC}$:

So, $\text{AH} = \frac{3}{5} \text{AG}$.

We know that $\text{AG} = \text{AH} + \text{HG}$.
Therefore, $\text{HG} = \text{AG} - \text{AH} = \text{AG} - \frac{3}{5} \text{AG} = \frac{2}{5} \text{AG}$.
The ratio $\text{AH} : \text{HG}$ is:

The ratio is $\boxed{3 : 2}$.

The parallelogram is $\text{DECF}$.
The base of the parallelogram is $\text{FC}$. From 10.1, $\text{FC} = \frac{3}{5} \text{BC}$.
Given $\text{BC} = (5-t)$ units.
So, $\text{FC} = \frac{3}{5}(5-t)$.
The perpendicular height of the parallelogram is $\text{HG}$. From 10.1, $\text{HG} = \frac{2}{5} \text{AG}$.
Given $\text{AG} = t$ units.
So, $\text{HG} = \frac{2}{5}t$.

The area of a parallelogram is given by base $\times$ perpendicular height.

To find the maximum area, we need to find the critical points by setting the first derivative to zero.

The second derivative is $\frac{d^2A}{dt^2} = \frac{6}{25}(-2) = -\frac{12}{25}$.
Since the second derivative is negative, $t = \frac{5}{2}$ corresponds to a maximum area.
Also, $t$ must be positive ($AG > 0$) and $5-t$ must be positive ($BC > 0$), so $0 < t < 5$. Our value $t = \frac{5}{2}$ is within this range.

The value of $t$ that will maximise the area of the parallelogram is $\boxed{\frac{5}{2}}$.