1961 AHSME Problems/Problem 8

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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:

(A) $C_1+C_2=A+B$

(B) $C_1-C_2=B-A$

(C) $C_1-C_2=A-B$

(D) $C_1+C_2=B-A$

(E) $C_1-C_2=A+B$

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