Track / Stage 5 / 167 of 400 #767 of 1964

Problem 767

AIME late

Geometry Difficulty 5.3 Find the answer

Given a triangle $ABC$ with $\measuredangle CBA=20^{\circ}, \measuredangle ACB=40^{\circ}$ and $\overline{AD}=2 \text{~cm}$ , where $D$ is the intersection point of the angle bisector of the angle at vertex $A$ and side $BC$ . Determine the difference $\overline{BC}-\overline{AB}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Solution. On side $BC$ , we choose a point $E$ such that $\measuredangle AEB=80^{\circ}$ . Then $\measuredangle BAE=80^{\circ}$ , so triangle $ABE$ is isosceles, i.e., $\overline{AB}=\overline{AE}$ . Triangle $AEC$ is isosceles because $\measuredangle EAC=40^{\circ}$ , so $\overline{AE}=\overline{CE}$ .

Triangle $ADE$ is isosceles because $\measuredangle ADE=80^{\circ}$ , so $\overline{AE}=\overline{AD}=$ $2 \text{ cm}$ . Therefore, $\overline{BC}-\overline{AB}=\overline{BC}-\overline{BE}=\overline{CE}=\overline{AE}=\overline{AD}=2 \text{ cm}$

## VII Section

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.