In a triangle with sides of lengths $a$, $b$, and $c$, $(a+b+c)(a+b-c) = 3ab$. The measure of the angle opposite the side length $c$ is
$\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ$
We will try to solve for a possible value of the variables. First notice that exchanging $a$ for $b$ in the original equation must also work. Therefore, $a=b$ works. Replacing $b$ for $a$ and expanding/simplifying in the original equation yields $4a^2-c^2=3a^2$, or $a^2=c^2$. Since $a$ and $c$ are positive, $a=c$. Therefore, we have an equilateral triangle and the angle opposite $c$ is just $\textbf{(D)}\ 60^\circ\qquad$.