# Difference between revisions of "1961 AHSME Problems/Problem 8"

## Problem

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:

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

## Solution

$[asy] draw((0,0)--(120,0)--(40,50)--(0,0)); draw((40,50)--(40,0)); draw((40,4)--(44,4)--(44,0)); label("B",(10,5)); label("A",(100,5)); label("C_1",(47,38)); label("C_2",(34,33)); [/asy]$

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 $\boxed{\textbf{(B)}}$.