AHSME 1961 Problem 8

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: $\textbf{(A)}\ C_1+C_2=A+B \textbf{(B)}\ C_1-C_2=B-A \\ \textbf{(C)}\ C_1-C_2=A-B \textbf{(D)}\ C_1+C_2=B-A \textbf{(E)}\ C_1-C_2=A+B$