Proof involving an isosceles triangle

Question

I came across this problem in some (maybe) high school book:

Let $ABC$ be an isosceles triangle s.t. $AB=AC$. Also, $\alpha>\beta$. It is known/given:

1. $\frac{BD}{DC}=\frac{\sin(\alpha)}{\sin(\beta)}$.
2. $\frac{S_{ABD}}{S_{ADC}}=\tan(\alpha)$.

Find the base angles of $\triangle ABC$.

I've tried pretty much everything involving the law of cosines/sines, and also auxiliary constructions of the normal to $BC$ (in $\triangle ABC$), which enables looking at the circumscribed circles of the two halves of $\triangle ABC$.

I will be glad to hear any insight about this problem. I think I'm missing something very elementary, as I didn't find the second equation too helpful. (The first one is obviously true for all isosceles triangles.)

Answer 1

Hint: It is indeed elementary. You are in effect told that $\frac{\sin\alpha}{\sin \beta}=\frac{\sin \alpha}{\cos\alpha}$. What does that tell you about the relationship between $\alpha$ and $\beta$?