<s>[INST] What is the area of the triangle ABC? 
$ABC$ is an equaliteral triangle.
Suppose $DB=4$, $DA=4\sqrt{3}$ and $DC=8$.
Find the area of the triangle $ABC$.
 [/INST] Reflect $D$ in $AB$ to obtain $P$. Reflect $D$ in $BC$ to obtain $Q$. Reflect $D$ in $CA$ ro obtain $R$. Then $[APBQCR]=2[\triangle ABC]$.
Note that $\triangle APR$, $\triangle BQP$ and $\triangle CRQ$ are $30^\circ$- $120^\circ$-$30^\circ$ isosceles triangles.
$[\triangle APR]=\frac{1}{2}(4\sqrt{3})^2\sin 120^\circ $
$[\triangle BQP]=\frac{1}{2}(4)^2\sin 120^\circ $
$[\triangle CRQ]=\frac{1}{2}(8)^2\sin 120^\circ$
By cosine formula, $PQ=4\sqrt{3}$, $PR=(4\sqrt{3})\sqrt{3}=12$ and $QR=8\sqrt{3}$.  It is easy to see that $\triangle PQR$ is right-angled and its area is easy to find.
Summing up, we will have the area of $APBQCR$.
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