Library / Macedonian Mathematical Competitions /14 of 73

Macedonian Mathematical Competitions

Geometry Difficulty 4.5 AIME Prove it North Macedonia

In the triangle $ABC$ , $\angle BAC = 120^\circ$ . On the bisector of the angle $\angle BAC$ , a point $D$ is chosen such that $\overline{AD} = \overline{AB} + \overline{AC}$ . Prove that $\triangle BCD$ is equilateral.

Solution

Let $M$ be a point on $AD$ , such that $\overline{AM} = \overline{AC}$ . $\angle MAC = 60^\circ$ implies that $\triangle ACM$ is an equilateral triangle. Because $\angle CAB = \angle CMD = 120^\circ$ , $\overline{MC} = \overline{AC}$ and $\overline{AB} = \overline{MD}$ we have that $\triangle CAB \cong \triangle CMD$ . Hence $\overline{BC} = \overline{CD}$ and $\angle ACB = \angle MCD$ . From $\angle ACM = \angle ACB + \angle BCM = \angle MCD + \angle BCM = \angle BCD = 60^\circ$ and $\overline{BC} = \overline{CD}$ we conclude that $\triangle BCD$ is an equilateral triangle.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.