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Let $\theta = \angle CAD, \phi = \angle CDB, \varphi=\angle DBC, \alpha = \angle BCD$ and $\beta=\angle ACD$. Then we have the following system of equations

$\theta + \varphi = 90^{\circ},$

$\theta + \beta = 120^{\circ},$

$\varphi + \alpha = 60^{\circ},$

$\alpha+\beta= 90^{\circ}.$

The associated matrix turns out to be invertible. Hence we can find all the angles and we can find $BC$.

But then my teacher told me there is a neater solution, I would like to ask is there any alternative and shorter solution, thank you so much.

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Use the first two equations to eliminate $\theta$, use the last two equations to eliminate $\alpha$, then you have two equations in the two unknowns $\phi,\beta$.

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  • $\begingroup$ After the (two) eliminations, there is only one equation left. $\endgroup$ – Mick Jul 11 '15 at 15:45
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If you are familiar with law of cosines in a triangle, then you can easily get length of $CD$ from a construction step. Let $O$ be the centre of circle with $AB$ as diameter. Then, $OC=2$. use Cosine law in $\triangle OCD$,

$$ cos(120)=\frac{ 1^2+CD^2-4}{2CD}\\ CD^2+CD-3=0 \\ CD=\frac{-1+\sqrt{13}}{2} $$ Now, In $\triangle BCD$,

$$ cos(120)=\frac{ 3^2+CD^2-BC^2}{2CD*3}\\ BC^2 = CD^2+3CD+9=3+2CD+9 =12+2CD\\ BC^2 =11+\sqrt{13}\\ BC=(11+\sqrt{13})^{1/2} $$

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