在下图所示的三角形中，一个顶点为原点 O，另外两个顶点为 A 和 B，其中 $\overrightarrow{OA}=\mathbf{a}$，$\overrightarrow{OB}=\mathbf{b}$。三角形的中线是连接顶点（O、A 或 B）与对边中点的线段（例如，AD 是三角形 OAB 从顶点 A 出发的中线）。已知 F 是三角形 OAB 从顶点 A 和 B 出发的中线的交点。

1. Find $\overrightarrow{AD}$ and $\overrightarrow{BE}$ using vector addition and the given midpoints.

Given D and E are the midpoints of OB and OA respectively:
$\overrightarrow{OD} = \frac{1}{2}\mathbf{b} \quad \text{and} \quad \overrightarrow{OE} = \frac{1}{2}\mathbf{a}$
Using vector subtraction:
$\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = \frac{1}{2}\mathbf{b} - \mathbf{a}$
$\overrightarrow{BE} = \overrightarrow{OE} - \overrightarrow{OB} = \frac{1}{2}\mathbf{a} - \mathbf{b}$

2. Express $\overrightarrow{OF}$ in two different ways using unknown scalars, as F lies on both AD and BE.

Since F lies on the line segment AD, points A, F, and D are collinear. Hence, $\overrightarrow{AF} = \lambda\overrightarrow{AD}$ for some scalar $\lambda$.
$\overrightarrow{OF} = \overrightarrow{OA} + \overrightarrow{AF} = \mathbf{a} + \lambda\left(\frac{1}{2}\mathbf{b} - \mathbf{a}\right)$
$\overrightarrow{OF} = (1 - \lambda)\mathbf{a} + \frac{1}{2}\lambda\mathbf{b}$

Similarly, F lies on BE, so points B, F, and E are collinear. Thus, $\overrightarrow{BF} = \mu\overrightarrow{BE}$ for some scalar $\mu$.
$\overrightarrow{OF} = \overrightarrow{OB} + \overrightarrow{BF} = \mathbf{b} + \mu\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right)$
$\overrightarrow{OF} = \frac{1}{2}\mu\mathbf{a} + (1 - \mu)\mathbf{b}$

3. Equate the coefficients of $\mathbf{a}$ and $\mathbf{b}$ from the two expressions to form linear equations.

Since $\mathbf{a}$ and $\mathbf{b}$ are non-zero, non-parallel vectors, we can equate their respective coefficients from the two expressions for $\overrightarrow{OF}$:
For $\mathbf{a}$: $1 - \lambda = \frac{1}{2}\mu$
For $\mathbf{b}$: $\frac{1}{2}\lambda = 1 - \mu$

4. Solve for the scalar and substitute it back to show the final expression for $\overrightarrow{OF}$.

From the second equation, we get $\mu = 1 - \frac{1}{2}\lambda$.
Substitute this into the first equation:
$1 - \lambda = \frac{1}{2}\left(1 - \frac{1}{2}\lambda\right)$
$1 - \lambda = \frac{1}{2} - \frac{1}{4}\lambda$
$\frac{1}{2} = \frac{3}{4}\lambda \implies \lambda = \frac{2}{3}$
Substitute $\lambda = \frac{2}{3}$ back into the expression for $\overrightarrow{OF}$:
$\overrightarrow{OF} = \left(1 - \frac{2}{3}\right)\mathbf{a} + \frac{1}{2}\left(\frac{2}{3}\right)\mathbf{b}$
$\overrightarrow{OF} = \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} = \frac{1}{3}(\mathbf{a} + \mathbf{b}) \quad \text{(shown)}$

1. Find the position vector of C, the midpoint of AB.

Since $OC$ is the median from vertex $O$, $C$ must be the midpoint of the opposite side $AB$.
$\overrightarrow{OC} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}$
$\overrightarrow{OC} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$

2. Show that $\overrightarrow{OF}$ is a scalar multiple of $\overrightarrow{OC}$ to prove collinearity.

From part (i), we have $\overrightarrow{OF} = \frac{1}{3}(\mathbf{a} + \mathbf{b})$.
Expressing $\overrightarrow{OF}$ in terms of $\overrightarrow{OC}$:
$\overrightarrow{OF} = \frac{2}{3} \left[ \frac{1}{2}(\mathbf{a} + \mathbf{b}) \right]$
$\overrightarrow{OF} = \frac{2}{3} \overrightarrow{OC}$
Since $\overrightarrow{OF}$ is a scalar multiple of $\overrightarrow{OC}$ and they share the common point $O$, the points $O$, $F$, and $C$ are collinear. Thus, $F$ lies on the line segment $OC$.