该图展示了一个以 $O$ 为圆心、 $AB$ 为直径的单位圆。点 $C$ 落在圆周上。 $A, B$ 和 $C$ 的位置向量分别为 $\mathbf{a}, \mathbf{b}$ 和 $\mathbf{c}$。

1. Express the vectors $\overrightarrow{AC}$ and $\overrightarrow{BC}$ in terms of position vectors $\mathbf{a}$ and $\mathbf{c}$.

Since $AB$ is a diameter with centre $O$, $O$ is the midpoint of $AB$. Therefore, $\overrightarrow{OB} = -\overrightarrow{OA}$, which means $\mathbf{b} = -\mathbf{a}$.

We can find the displacement vectors:
$\overrightarrow{AC} = \mathbf{c} - \mathbf{a}$
$\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = \mathbf{c} - (-\mathbf{a}) = \mathbf{c} + \mathbf{a}$

2. Evaluate the scalar product of $\overrightarrow{AC}$ and $\overrightarrow{BC}$.

$\overrightarrow{AC} \cdot \overrightarrow{BC} = (\mathbf{c} - \mathbf{a}) \cdot (\mathbf{c} + \mathbf{a})$
Expanding the scalar product:
$\overrightarrow{AC} \cdot \overrightarrow{BC} = \mathbf{c} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{a}$
Since the scalar product is commutative ($\mathbf{c} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{c}$) and $\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^2$:
$\overrightarrow{AC} \cdot \overrightarrow{BC} = |\mathbf{c}|^2 - |\mathbf{a}|^2$

3. Apply the properties of a unit circle to conclude perpendicularity.

Since $A$ and $C$ lie on a unit circle with centre $O$, their distance from the origin is the radius of the circle, which is $1$.
Thus, $|\mathbf{a}| = 1$ and $|\mathbf{c}| = 1$.

$\overrightarrow{AC} \cdot \overrightarrow{BC} = (1)^2 - (1)^2 = 0$

Since the scalar product of $\overrightarrow{AC}$ and $\overrightarrow{BC}$ is $0$, the vectors are perpendicular. Hence, $AC$ and $BC$ are perpendicular.

1. Identify the vectors and the magnitude of $\mathbf{a}$.

The vector $\mathbf{c} - \mathbf{a} = \overrightarrow{AC}$ and $\mathbf{a} = \overrightarrow{OA}$.
Since the circle is a unit circle, the radius is $1$, so $|\mathbf{a}| = 1$.

2. Interpret the dot product geometrically.

$|\mathbf{a} \cdot (\mathbf{c} - \mathbf{a})| = |\overrightarrow{OA} \cdot \overrightarrow{AC}|$
Since $|\overrightarrow{OA}| = 1$, we can write this as $\frac{|\overrightarrow{OA} \cdot \overrightarrow{AC}|}{|\overrightarrow{OA}|}$.
The absolute value of the scalar product of a vector with a unit vector gives the length of the projection of that vector onto the direction of the unit vector.

1. Find the position vectors of $B$ and $R$ in terms of $\mathbf{a}$.

Since $O$ is the centre and $AB$ is the diameter, $\mathbf{b} = -\mathbf{a}$.
As $R$ lies on $OA$ produced with $OA:AR = 1:2$, we have $\overrightarrow{OR} = 3\overrightarrow{OA}$. Thus, the position vector of $R$ is $\mathbf{r} = 3\mathbf{a}$.

2. Find the position vectors of $D$ and $M$ in terms of $\mathbf{a}$ and $\mathbf{c}$.

Since $ABCD$ is a parallelogram, opposite sides are equal and parallel, so $\overrightarrow{AD} = \overrightarrow{BC}$.
$\mathbf{d} - \mathbf{a} = \mathbf{c} - \mathbf{b}$
$\mathbf{d} = \mathbf{a} + \mathbf{c} - (-\mathbf{a}) = 2\mathbf{a} + \mathbf{c}$
Since $M$ is the mid-point of $AD$, its position vector is:
$\mathbf{m} = \frac{\mathbf{a} + \mathbf{d}}{2} = \frac{\mathbf{a} + 2\mathbf{a} + \mathbf{c}}{2} = \frac{3\mathbf{a} + \mathbf{c}}{2}$

3. Evaluate $\overrightarrow{CM} \times \overrightarrow{CR}$ to show collinearity.

Find the vectors $\overrightarrow{CM}$ and $\overrightarrow{CR}$:
$\overrightarrow{CM} = \mathbf{m} - \mathbf{c} = \frac{3\mathbf{a} + \mathbf{c}}{2} - \mathbf{c} = \frac{1}{2}(3\mathbf{a} - \mathbf{c})$
$\overrightarrow{CR} = \mathbf{r} - \mathbf{c} = 3\mathbf{a} - \mathbf{c}$
Notice that $\overrightarrow{CR} = 2\overrightarrow{CM}$. Now, evaluate the required cross product:
$\overrightarrow{CM} \times \overrightarrow{CR} = \overrightarrow{CM} \times (2\overrightarrow{CM}) = 2(\overrightarrow{CM} \times \overrightarrow{CM})$
Since the cross product of any vector with itself is the zero vector ($\mathbf{v} \times \mathbf{v} = \mathbf{0}$), we have:
$\overrightarrow{CM} \times \overrightarrow{CR} = \mathbf{0}$
This implies that $\overrightarrow{CM}$ and $\overrightarrow{CR}$ are parallel. Since $C$ is a common point for both vectors, the points $C$, $M$ and $R$ are collinear.