Singapur

Geometry Difficulty 4.4 AIME Prove it Singapore

In the triangle $ABC$ , $\angle A = 90^\circ$ , the bisector of $\angle B$ meets the altitude $AD$ at the point $E$ , and the bisector of $\angle CAD$ meets the side $CD$ at $F$ . The line through $F$ perpendicular to $BC$ intersects $AC$ at $G$ . Prove that $B, E, G$ are collinear.

Solution

First $\triangle ABC \sim \triangle DAC$ so that $AC/BA = DC/AD$ . Also $\triangle DAC \sim \triangle DBA$ . It follows that $\triangle AFC \sim \triangle BEA$ so that $FC/AC = EA/BA$ . Thus $FC/EA = AC/BA = DC/AD$ . This shows that $EF$ is parallel to $AC$ . Hence $AEFG$ is a parallelogram. Since $\angle EFA = \angle FAG = \angle FAE$ , we have $AEFG$ is a rhombus so that $AG = FG$ . Thus $\triangle BFG \cong \triangle BAG$ . Consequently, $\angle FBG = \angle ABG = \frac{1}{2}\angle ABC = \angle FBE$ . This means $B, E, G$ are collinear.

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