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 $ABC$.
 A: HINT
Use Heron's formula for the three triangles $DBC,DAC,DAB$ and equate to the area for $\triangle ABC$ to find the unknown side $L$.
$$\sqrt{s_1(s_1-4)(s_1-4\sqrt 3)(s_1-L)}+\sqrt{s_2(s_2-4)(s_2-L)(s_2-8)}+\sqrt{s_3(s_3-L)(s_3-4\sqrt 3)(s_3-8)}=\frac{\sqrt 3}4L^2$$
with


*

*$s_1=\frac12\left(4+4\sqrt 3+L\right)$

*$s_2=\frac12\left(4+8+L\right)$

*$s_3=\frac12\left(4\sqrt 3+8+L\right)$

A: 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 $APBQCR$.
A: HINT: use that $$a^2=16+48-32\sqrt{3}\cos(\alpha),a^2=16+64-64\cos(\beta),a^2=64+48-64\sqrt{3}\cos(\gamma),\alpha+\beta+\gamma=360^{\circ}$$
you have four equations with the unknowns $$\alpha,\beta,\gamma,a$$
A: We'll rotate $\Delta ADC$ on $-60^{\circ}$ around the point $C$.
Let $R_{C}^{-90^{\circ}}$ this rotation and $R_{C}^{-90^{\circ}}(D)=D_1.$
Thus, $$R_{C}^{-90^{\circ}}(A)=B$$ and since $\Delta DD_1C$ is an equilateral triangle, we obtain $$DD_1=DC=8,$$ 
$$BD_1=4\sqrt3,$$ which gives $$\measuredangle DBD_1=90^{\circ},$$ $$\measuredangle BD_1D=30^{\circ}$$ and
$$\measuredangle BD_1C=30^{\circ}+60^{\circ}=90^{\circ},$$
which gives
$$BC^2=(4\sqrt3)^2+8^2=112$$ and
$$S_{\Delta ABC}=\frac{BC^2\sqrt3}{4}=28\sqrt3.$$
