Point $P$ is taken interior to a square with side-length $a$ and such that is it equally distant from two 
consecutive vertices and from the side opposite these vertices. If $d$ represents the common distance, then $d$ equals:
$\textbf{(A)}\ \frac{3a}{5}\qquad \textbf{(B)}\ \frac{5a}{8}\qquad \textbf{(C)}\ \frac{3a}{8}\qquad \textbf{(D)}\ \frac{a\sqrt{2}}{2}\qquad \textbf{(E)}\ \frac{a}{2}$
Draw a diagram and label it as shown.  Because of SSS Congruency, the two bottom triangles are [right triangles](https://artofproblemsolving.com/wiki/index.php/Right_triangle).  By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem),
\[d^2 = (a-d)^2 + (\frac{a}{2})^2\]
\[d^2 = a^2 - 2ad + d^2 + \frac{a^2}{4}\]
\[2ad = \frac{5a^2}{4}\]
\[d = \frac{5a}{8}\]
Thus, the answer is $\boxed{\textbf{(B)}}$.