以原点 $O$ 为参考，点 $A$ 和 $B$ 的位置向量分别为 $\mathbf{a}$ 和 $\mathbf{b}$，其中 $\mathbf{a}$ 和 $\mathbf{b}$ 是非零且不平行的向量。点 $C$ 在 $OA$ 上，使得 $OC : CA = 2 : 1$。点 $D$ 在 $OB$ 的延长线上，使得 $OD : BD = 4 : 1$。

1. Find the position vector of $C$.

Since $C$ lies on $OA$ such that $OC : CA = 2 : 1$, the length of $OC$ is $\frac{2}{2+1} = \frac{2}{3}$ of the total length of $OA$.
$\overrightarrow{OC} = \frac{2}{3}\overrightarrow{OA} = \frac{2}{3}\mathbf{a}$

2. Find the position vector of $D$.

Since $D$ lies on $OB$ produced, the points are collinear in the order $O, B, D$.
Given $OD : BD = 4 : 1$, the length $OB = OD - BD$ corresponds to $4 - 1 = 3$ parts.
Thus, the ratio $OD : OB = 4 : 3$.
$\overrightarrow{OD} = \frac{4}{3}\overrightarrow{OB} = \frac{4}{3}\mathbf{b}$

1. State the position vectors of $C$ and $D$ from part (i), and formulate an expression for $\overrightarrow{OE}$ using the line $BC$.

From part (i), we have $\overrightarrow{OC} = \frac{2}{3}\mathbf{a}$ and $\overrightarrow{OD} = \frac{4}{3}\mathbf{b}$.

Since $E$ lies on the line passing through $B$ and $C$, its position vector can be expressed using a parameter $\lambda$:
$\overrightarrow{OE} = (1 - \lambda)\overrightarrow{OC} + \lambda \overrightarrow{OB}$
$\overrightarrow{OE} = \frac{2}{3}(1 - \lambda)\mathbf{a} + \lambda \mathbf{b}$

2. Formulate a second expression for $\overrightarrow{OE}$ using the line $AD$.

Similarly, since $E$ lies on the line passing through $A$ and $D$, we can express its position vector using another parameter $\mu$:
$\overrightarrow{OE} = (1 - \mu)\overrightarrow{OA} + \mu \overrightarrow{OD}$
$\overrightarrow{OE} = (1 - \mu)\mathbf{a} + \frac{4}{3}\mu \mathbf{b}$

3. Equate the two expressions and compare the coefficients of the non-parallel vectors $\mathbf{a}$ and $\mathbf{b}$.

Equating the two expressions for $\overrightarrow{OE}$:
$\frac{2}{3}(1 - \lambda)\mathbf{a} + \lambda \mathbf{b} = (1 - \mu)\mathbf{a} + \frac{4}{3}\mu \mathbf{b}$

Since $\mathbf{a}$ and $\mathbf{b}$ are non-zero and non-parallel vectors, their respective coefficients must be equal.
For $\mathbf{a}$:
$\frac{2}{3}(1 - \lambda) = 1 - \mu \quad \text{--- (1)}$
For $\mathbf{b}$:
$\lambda = \frac{4}{3}\mu \quad \text{--- (2)}$

4. Solve the simultaneous linear equations to find the parameter $\mu$.

Substitute equation (2) into equation (1):
$\frac{2}{3}\left(1 - \frac{4}{3}\mu\right) = 1 - \mu$
Multiply both sides by $3$:
$2\left(1 - \frac{4}{3}\mu\right) = 3 - 3\mu$
$2 - \frac{8}{3}\mu = 3 - 3\mu$
Multiply by $3$ again to clear the denominator:
$6 - 8\mu = 9 - 9\mu$
$9\mu - 8\mu = 9 - 6$
$\mu = 3$

5. Substitute the parameter back into the expression for $\overrightarrow{OE}$ to show the required position vector.

Substitute $\mu = 3$ into the expression for $\overrightarrow{OE}$ derived from the line $AD$:
$\overrightarrow{OE} = (1 - 3)\mathbf{a} + \frac{4}{3}(3)\mathbf{b}$
$\overrightarrow{OE} = -2\mathbf{a} + 4\mathbf{b}$
$\overrightarrow{OE} = 4\mathbf{b} - 2\mathbf{a} \quad \text{(Shown)}$

1. Find the displacement vectors $\overrightarrow{CD}$ and $\overrightarrow{CE}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

From earlier parts, we know $\overrightarrow{OC} = \frac{2}{3}\mathbf{a}$, $\overrightarrow{OD} = \frac{4}{3}\mathbf{b}$, and $\overrightarrow{OE} = 4\mathbf{b} - 2\mathbf{a}$.
$\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = \frac{4}{3}\mathbf{b} - \frac{2}{3}\mathbf{a}$
$\overrightarrow{CE} = \overrightarrow{OE} - \overrightarrow{OC} = (4\mathbf{b} - 2\mathbf{a}) - \frac{2}{3}\mathbf{a} = 4\mathbf{b} - \frac{8}{3}\mathbf{a}$

2. Set up the area of the triangle using the vector product and expand the expression.

The area of $\triangle CDE$ is given by $\frac{1}{2}|\overrightarrow{CD} \times \overrightarrow{CE}|$.
$\overrightarrow{CD} \times \overrightarrow{CE} = \left(\frac{4}{3}\mathbf{b} - \frac{2}{3}\mathbf{a}\right) \times \left(4\mathbf{b} - \frac{8}{3}\mathbf{a}\right)$
Using the distributive property and noting that $\mathbf{a} \times \mathbf{a} = \mathbf{0}$ and $\mathbf{b} \times \mathbf{b} = \mathbf{0}$:
$\overrightarrow{CD} \times \overrightarrow{CE} = \frac{4}{3}\mathbf{b} \times \left(-\frac{8}{3}\mathbf{a}\right) - \frac{2}{3}\mathbf{a} \times 4\mathbf{b}$
$= -\frac{32}{9}(\mathbf{b} \times \mathbf{a}) - \frac{8}{3}(\mathbf{a} \times \mathbf{b})$

3. Simplify using the anti-commutative property of the cross product to evaluate $k$.

Since $\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})$:
$\overrightarrow{CD} \times \overrightarrow{CE} = \frac{32}{9}(\mathbf{a} \times \mathbf{b}) - \frac{24}{9}(\mathbf{a} \times \mathbf{b}) = \frac{8}{9}(\mathbf{a} \times \mathbf{b})$
$\text{Area of } \triangle CDE = \frac{1}{2} \left| \frac{8}{9}(\mathbf{a} \times \mathbf{b}) \right| = \frac{4}{9}|\mathbf{a} \times \mathbf{b}|$
Comparing this to $k|\mathbf{a} \times \mathbf{b}|$, we find $k = \frac{4}{9}$.