MathNet 2021 — Part 04 Question 63

In $\triangle ABC$, $AC = kAB$, with $k \gt 1$. 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 $k$.