Why does the concentration of the acid equals that of the base at the mid-point of titration?
At the midpoint of titration, the concentration of acid is just approximately equal to its conjugated base, as there must be considered the acid dissociation and ( in less extend ) salt hydrolysis.

If we take the definition(simplified using concentrations):

$$K_\mathrm{a}=\frac {\ce{[H+][A-]}}{\ce{[HA]}}$$

we can see that by definition:

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log { \frac {\ce{[A-]}}{\ce{[HA]}}}$$

$\mathrm{p}K_\mathrm{a}$ is $\mathrm{pH}$ when both items of the conjugate pair have the same concentration.

$\mathrm{pH}$ near titration midpoint is approximately equal

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log { \frac {n_\mathrm{base}}{
n_\mathrm{acid, init} - n_\mathrm{base}
}}$$

This function has minimal slope = maximal buffer capacity at $$n_\mathrm{acid, init} = 2 \cdot n_\mathrm{base}$$