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Geometry Difficulty 4.8 AIME Prove it Hong Kong

In $\triangle ABC$ , $AC = kAB$ , with $k > 1$ . The internal angle bisector of $\angle BAC$ meets $BC$ at $D$ . The circle with $AC$ as diameter cuts the extension of $AD$ at $E$ . Express $\frac{AD}{AE}$ in terms of $k$ .

Solution

The answer is $\frac{2}{k+1}$ .

Let $CE$ meet $AB$ at $Q$ . Let $P$ be the point on $AQ$ such that $PD // QC$ . Note that $E$ is the midpoint of $QC$ since $\triangle AEQ \cong \triangle AEC$ . Also, we have $\triangle APD \sim \triangle AQE$ and $\triangle BPD \sim \triangle BQC$ . Thus, we have
$\frac{AD}{AE} = \frac{PD}{QE} = 2 \cdot \frac{PD}{QC} = 2 \cdot \frac{BD}{BC}.$
By the angle bisector theorem, $\frac{DB}{DC} = \frac{AB}{AC} = \frac{1}{k}$ . It follows that $\frac{AD}{AE} = \frac{2}{k+1}$ .

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