Singapur

Geometry Difficulty 4.4 AIME Prove it Singapore

In a triangle $ABC$ , $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$ . The line through $B$ perpendicular to $AD$ intersects $AD$ at $E$ and $AM$ at $G$ . Prove that $GD$ is parallel to $AB$ .

Solution

Let $BE$ intersect $AC$ at $H$ . Thus $AE$ is the perpendicular bisector of $BH$ and so $ME$ is parallel to $CA$ and $\frac{HC}{ME} = 2$ . Hence the triangles $MEG$ and $AHE$ are similar. Also the triangles $DME$ and $DCA$ are similar. It follows that $\frac{AG}{GM} = \frac{AH}{ME} = \frac{AC - HC}{ME} = \frac{AC}{ME} - \frac{HC}{ME} = \frac{CD}{DM} - 2 = \frac{MC + DM}{DM} - 2 = \frac{BM + DM}{DM} - 2 = \frac{BM}{DM} - 1 = \frac{BM - DM}{DM} = \frac{BD}{DM}$ . Therefore, $GD$ is parallel to $AB$ .

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