Analysis

This question examines the operation of the scalar product of plane vectors and the calculation of vector projection, which is a basic problem.

According to the definition of "congruence", write out $\dfrac{|\overrightarrow{a}|}{|\overrightarrow{b}|} =\cos \theta$, then calculate the scalar product $(\overrightarrow{a}-\overrightarrow{b})\cdot \overrightarrow{a}$, thereby finding the projection of $\overrightarrow{a}-\overrightarrow{b}$ on $\overrightarrow{a}$.

Solution

According to the problem, $\dfrac{|\overrightarrow{a}|}{|\overrightarrow{b}|} =\cos \theta$, where $\theta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$;

Therefore, $(\overrightarrow{a}-\overrightarrow{b})\cdot \overrightarrow{a} = \overrightarrow{a}^{2}-\overrightarrow{a}\cdot \overrightarrow{b} = \overrightarrow{a}^{2}-|\overrightarrow{a}|\cdot|\overrightarrow{b}|\cdot\dfrac{|\overrightarrow{a}|}{|\overrightarrow{b}|} = \overrightarrow{a}^{2}-\overrightarrow{b}^{2}$;
Thus, the projection of $\overrightarrow{a}-\overrightarrow{b}$ on $\overrightarrow{a}$ is:
$|\overrightarrow{a}-\overrightarrow{b}|\cos$

$=|\overrightarrow{a}-\overrightarrow{b}|\times\dfrac{(\overrightarrow{a}-\overrightarrow{b})\cdot \overrightarrow{a}}{|\overrightarrow{a}-\overrightarrow{b}|\times |\overrightarrow{a}|}$

$=\dfrac{\overrightarrow{a}^{2}-\overrightarrow{b}^{2}}{|\overrightarrow{a}|}$.
Therefore, the correct option is $\boxed{D}$.