Library / Harvard-MIT Mathematics Tournament /79 of 352

Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.8 AIME Prove it United States

Problem:
Let $ABC$ be a triangle with $AB < AC$ . Let the angle bisector of $\angle A$ and the perpendicular bisector of $BC$ intersect at $D$ . Then let $E$ and $F$ be points on $AB$ and $AC$ such that $DE$ and $DF$ are perpendicular to $AB$ and $AC$ , respectively. Prove that $BE = CF$ .

Solution

Solution:
Note that $DE$ , $DF$ are the distances from $D$ to $AB$ , $AC$ , respectively, and because $AD$ is the angle bisector of $\angle BAC$ , we have $DE = DF$ . Also, $DB = DC$ because $D$ is on the perpendicular bisector of $BC$ . Finally, $\angle DEB = \angle DFC = 90^{\circ}$ , so it follows that $\triangle DEB \cong \triangle DFC$ , and $BE = CF$ .

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.