# Would like some pointers on this geometry problem

I'm working on this problem (not homework):

Let $ABC$ be a triangle and $D$ a point on the side $AC$. If $C\hat{B}D-A\hat{B}D = 60°$, $B\hat{D}C = 30°$ and $AB \cdot BC =BD^2$, find the angles of the triangle $ABC$.

I've been working on it for a while, and have managed to express all the angles in terms of a single variable, but I have no idea how to use the relationship between the sides. I've tried to construct right triangles to use trigonometry or Pythagoras' Theorem, but it just ends up introducing new sides I don't care about. I drew a picture to make things clearer for me, but sadly I don't have a scanner or a camera right now, so I can't upload it.

It would be great if you could give me a few tips instead of the full answer; I'd like to figure it out (mostly) on my own.

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Drop a perpendicular from $B$ to the side $AC$ and obtain $E$. You may assume $BE=1$. The triangle $DBE$ is a $30^\circ/60^\circ/90^\circ$ triangle, whence $BD=2$. Let $\beta:=\angle(ABD)=\angle(EBC)\$. Then $BC$ and $BA$ can be expressed in terms of $\beta$, resp., $\beta+60^\circ$, and the condition $AB\cdot BC = BD^2$ amounts to $$\cos\beta\cdot \cos(\beta+60^\circ)={1\over4}\ .$$ Plot the left side of this equation as a function of $\beta$, and you will get a conjecture about $\beta$ which is easy to prove.

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Sorry, it's $AB \cdot BC = BD^2$, the blockquote messed it up. But the idea of assigning some random value to a side might be helpful. – Javier Apr 5 '12 at 15:19
Also, how did you get $\cos \beta \cdot \cos (\beta + 60°) \le \frac1{4}$? – Javier Apr 5 '12 at 15:22
This was helpful, thank you! – Javier Apr 5 '12 at 18:50

$$\cos\beta\cdot \cos(\beta+60^\circ)={1\over4}\ .$$

implies

$$2\cos\beta\cdot \cos(\beta+60^\circ)={1\over2}\ .$$

so

$$\cos60^\circ + \cos(2\cdot\beta+60^\circ)={1\over2}\ .$$

so

$$\cos(2\cdot\beta+60^\circ)=0\ .$$

So $$(2\cdot\beta+60^\circ) = 90^\circ$$

Thus $\beta$ is $15^\circ$.

Thus in ABC, angles A, B and C are 15, 90 and 75 degrees respectively.

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