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

Isosceles triangle

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.)

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  • $\begingroup$ What does $S_{ABD}$ mean? $\endgroup$
    – André Nicolas
    Aug 5, 2014 at 6:20
  • $\begingroup$ $\triangle ABD$'s area. (The ratio easily simplifies, but I wanted to give the original equation in case I missed something.) $\endgroup$
    – Isosceles
    Aug 5, 2014 at 6:34

1 Answer 1

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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 $\beta$?

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  • $\begingroup$ $sin(\beta)=cos(\alpha)$, 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). $\endgroup$
    – Isosceles
    Aug 5, 2014 at 6:48
  • $\begingroup$ Much simpler than that. Because $\alpha\gt \beta$, we have $\alpha+\beta=90^\circ$. The end. $\endgroup$
    – André Nicolas
    Aug 5, 2014 at 6:52
  • $\begingroup$ Thank you, but how does it follow from the equation? (If possible, a hint rather than an answer.) $\endgroup$
    – Isosceles
    Aug 5, 2014 at 7:08
  • $\begingroup$ 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. $\endgroup$
    – André Nicolas
    Aug 5, 2014 at 7:14

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