The problem is as follows:

In triangle $ABC$, $BC=2$. Point $D$ is on $\overline{AC}$ such that $AD=1$ and $CD=2$. If $m\angle BDC=2m\angle A$, compute $\sin A$.

I tried several ways of making similar triangles, but that didn't work.I couldn't find a simple way to apply the double angle identities either, and I think that's probably where I messed up. Is there a good way to do this without the double angle identities?


Let be $\measuredangle BAC=\alpha$ and $\measuredangle BDC=2\alpha$. Since $CD=BC=2$, we have that $\measuredangle CBD=\measuredangle BDC=2\alpha$. Also,

\begin{eqnarray} \measuredangle BAD+\measuredangle ABD&=&\measuredangle BDC, \\ \measuredangle BAD+\alpha&=&2\alpha, \\ \measuredangle BAD&=&\alpha. \end{eqnarray}

We have $BD=AD=1$ (since $\measuredangle BAD=\measuredangle DAB$, in triangle $ABD$).

Now, from law of cosines (in triangle $BCD$), we get $2^2=2^2+1^2-2*1*2*\cos{2\alpha}$ and $\frac{1}{4}=\cos{2\alpha}=1-2\sin^2{\alpha}$.

Finally, $\sin{\alpha}=\sqrt{\frac{3}{8}}$.

  • $\begingroup$ Thank you. I had gotten the first para and those equations, I just didnt have time to apply law of cosines and work it out. $\endgroup$ – Asimov Jun 4 '14 at 20:56

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