# Proof involving an isosceles triangle

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.)

• What does $S_{ABD}$ mean? – André Nicolas Aug 5 '14 at 6:20
• $\triangle ABD$'s area. (The ratio easily simplifies, but I wanted to give the original equation in case I missed something.) – Isosceles Aug 5 '14 at 6:34

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 $\beta$?
• $sin(\beta)=cos(\alpha)$, but I don't see how it helps (I have repeatedly used this substitution in the equations I got from the law of cosines/sines, but nothing turned up). – Isosceles Aug 5 '14 at 6:48
• Much simpler than that. Because $\alpha\gt \beta$, we have $\alpha+\beta=90^\circ$. The end. – André Nicolas Aug 5 '14 at 6:52
• A hint is impossible, since it is so easy. The angle $\beta$ is under $90$ degrees. The angle $\alpha$ is also under $90$ degrees since its cosine is positive. Note that $\cos(\alpha)=\sin(90^\circ -\alpha)$. It follows that $90^\circ-\alpha=\beta$. So $\alpha+\beta=90^{\circ}$. Now the base angles are obvious. – André Nicolas Aug 5 '14 at 7:14