1960 AMC 12 Problem 38

Question

In this diagram

\overline{AB}

and

\overline{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:

b=\dfrac{a+c}{2}

b=\dfrac{a-c}{2}

a=\dfrac{b-c}{2}

a=\dfrac{b+c}{2}

none of these

Let each base angle of

ABC

be

\theta

Compare the directions of

DF,

DE,

and

EF

using the

60^\circ

angles of the equilateral triangle

Let each base angle of

ABC

be

\theta.

Measured from the direction

BC,

the line

FD

has direction

180^\circ-a,

so the equilateral condition makes

DE

have direction

60^\circ-a.

At

D,

b=\theta-(60^\circ-a)=\theta-60^\circ+a.

Similarly, angle chasing at

E

gives

c=60^\circ+a-\theta.

Adding these equations yields

b+c=2a.

Therefore

a=\frac{b+c}{2},

and the correct answer is D.

Problem 38 in other years