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 $MN$

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

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

Answer: C

Concepts:angle bisector theorem ratio and proportion parallel lines

Difficulty rating: 1770

Small Hint:

Use $BM=BD,\ CN=CD,$ and the angle bisector theorem

Big Hint:

Show that $\frac{BM}{CN}=\frac{AB}{AC}$ and apply the converse of the side-splitter theorem

Solution:

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.

1975 AMC 12 Problem 26

Answer: C

Difficulty rating: 1770

Small Hint:

Big Hint:

Solution:

Problem 26 in Other Years