Let the two base angles of a triangle be $A$ and $B$, with $B$ larger than $A$. 
The altitude to the base divides the vertex angle $C$ into two parts, $C_1$ and $C_2$, with $C_2$ adjacent to side $a$. Then:
$\textbf{(A)}\ C_1+C_2=A+B \qquad \textbf{(B)}\ C_1-C_2=B-A \qquad\\ \textbf{(C)}\ C_1-C_2=A-B \qquad \textbf{(D)}\ C_1+C_2=B-A\qquad \textbf{(E)}\ C_1-C_2=A+B$
Noting that side $a$ is opposite of angle $A$, label the diagram as shown.
From the two right triangles,
\[B + C_2 = 90\]
\[A + C_1 = 90\]
Substitute to get
\[B + C_2 = A + C_1\]
\[B - A = C_1 - C_2\]
The answer is $\boxed{\textbf{(B)}}$.