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

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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