Step 1: Identify the median vector CD.
$CD$ is the median from vertex $C$ to side $AB$, so $D$ is the midpoint of $AB$. By the midpoint formula in vectors, $\overrightarrow{CD} = \frac{1}{2}(\overrightarrow{CA}+\overrightarrow{CB})$. This is because $D$ divides $AB$ in the ratio $1:1$.
Step 2: Set up the dot product.
\[ \overrightarrow{CA}\cdot\overrightarrow{CD} = \overrightarrow{CA}\cdot\frac{1}{2}(\overrightarrow{CA}+\overrightarrow{CB}) = \frac{1}{2}\left(|\overrightarrow{CA}|^2 + \overrightarrow{CA}\cdot\overrightarrow{CB}\right). \]
Step 3: Identify the magnitudes.
We have $|\overrightarrow{CB}|=a$, $|\overrightarrow{CA}|=b$, $|\overrightarrow{AB}|=c$. So $|\overrightarrow{CA}|^2 = b^2$.
Step 4: Find CA dot CB using the law of cosines.
The angle at $C$ between $\overrightarrow{CA}$ and $\overrightarrow{CB}$ is angle $C$. By the cosine rule, $c^2 = a^2+b^2-2ab\cos C$, giving $\cos C = \frac{a^2+b^2-c^2}{2ab}$. So: \[ \overrightarrow{CA}\cdot\overrightarrow{CB} = |\overrightarrow{CA}||\overrightarrow{CB}|\cos C = b \cdot a \cdot \frac{a^2+b^2-c^2}{2ab} = \frac{a^2+b^2-c^2}{2}. \]
Step 5: Substitute everything in.
\[ \overrightarrow{CA}\cdot\overrightarrow{CD} = \frac{1}{2}\left(b^2 + \frac{a^2+b^2-c^2}{2}\right) = \frac{1}{2}\cdot\frac{2b^2+a^2+b^2-c^2}{2} = \frac{a^2+3b^2-c^2}{4}. \]
Step 6: State the final answer.
\[ \boxed{\dfrac{a^2+3b^2-c^2}{4}} \]