Isosceles Triangle and Ratio of Altitudes

In $\Delta ABC$, $AB = AC$, $AD \perp BC$, and $BE \perp AC$. $AD$ bisects $\angle A$ and $BC$.

In $\Delta AOB$, $\angle AOB = 105^\circ \implies \angle BOD = 180^\circ - 105^\circ = 75^\circ$. In right triangle $BOD$ (with $\angle BDO = 90^\circ$): $\angle OBD = 90^\circ - 75^\circ = 15^\circ$

So base angle $C = B = 90^\circ - 15^\circ = 75^\circ$. Then $\angle A = 180^\circ - 2(75^\circ) = 30^\circ$.

Comparing area representations of $\Delta ABC$: $\text{Area} = \frac{1}{2} \cdot BC \cdot AD = \frac{1}{2} \cdot AC \cdot BE \implies \frac{AD}{BE} = \frac{AC}{BC}$

By Sine Rule in $\Delta ABC$: $\frac{AC}{\sin B} = \frac{BC}{\sin A} \implies \frac{AC}{BC} = \frac{\sin 75^\circ}{\sin 30^\circ} = \frac{\cos 15^\circ}{1/2} = 2\cos 15^\circ$