# Difference between revisions of "1960 AHSME Problems/Problem 38"

## Problem

In this diagram $AB$ and $AC$ are the equal sides of an isosceles $\triangle ABC$, in which is inscribed equilateral $\triangle DEF$. Designate $\angle BFD$ by $a$, $\angle ADE$ by $b$, and $\angle FEC$ by $c$. Then:

$[asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); pair A=(5,12),B=origin,C=(10,0),D=(5/3,4),E=(10-5*.45,12*.45),F=(6,0); draw(A--B--C--cycle^^D--E--F--cycle); draw(anglemark(E,D,A,1,45)^^anglemark(F,E,C,1,45)^^anglemark(D,F,B,1,45)); label("b",(D.x+.2,D.y+.25),dir(30)); label("c",(E.x,E.y-.4),S); label("a",(F.x-.4,F.y+.1),dir(150)); label("A",A,N); label("B",B,S); label("C",C,S); label("D",D,dir(150)); label("E",E,dir(60)); label("F",F,S);[/asy]$

$\textbf{(A)}\ b=\frac{a+c}{2}\qquad \textbf{(B)}\ b=\frac{a-c}{2}\qquad \textbf{(C)}\ a=\frac{b-c}{2} \qquad \textbf{(D)}\ a=\frac{b+c}{2}\qquad \textbf{(E)}\ \text{none of these}$

## Solution

Since $\triangle DEF$ is an equilateral triangle, all of the angles are $60^{\circ}$. The angles in a line add up to $180^{\circ}$, so $$\angle FDB = 120 - b$$ $$\angle EFC = 120 - a$$ The angles in a triangle add up to $180^{\circ}$, so $$\angle ABC = 60 + b - a$$ $$\angle ACB = 60 - c + a$$ Since $\triangle ABC$ is isosceles and $AB = AC$, by Base-Angle Theorem, $$60 + b - a = 60 - c + a$$ $$b + c = 2a$$ $$a = \frac{b+c}{2}$$ The answer is $\boxed{\textbf{(D)}}$.