1975 AMC 12 Problem 26

Question

In acute triangle

ABC

the bisector of

\angle A

meets side

BC

at

D.

The circle with center

B

and radius

BD

intersects side

AB

at

M;

and the circle with center

C

and radius

CD

intersects side

AC

at

N.

Then it is always true that

\begin{aligned}\angle CND+\angle BMD\\{}-\angle DAC=120^\circ\end{aligned}

AMDN

is a trapezoid

BC

is parallel to

MN

AM-AN=\frac{3(DB-DC)}2

AB-AC=\frac{3(DB-DC)}2

Use

BM=BD,\ CN=CD,

and the angle bisector theorem

Show that

\frac{BM}{CN}=\frac{AB}{AC}

and apply the converse of the side-splitter theorem

The angle bisector theorem gives

\frac{BD}{CD}=\frac{AB}{AC}.

Since

BM=BD

and

CN=CD,

we have

\frac{BM}{CN}=\frac{AB}{AC}.

Therefore

M

and

N

divide

AB

and

AC

proportionally from

B

and

C,

so the converse of the side-splitter theorem gives

MN\parallel BC.

Therefore, the correct answer is C.

Problem 26 in other years