A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.


Consider an equilateral  triangle ABC and a Point P such that $\overline{AP} = 12$, $\overline{BP} = 13$ and $\overline{CP} = 5$.
Rotate $P$ clockwise along $A$ by $60^o$ to $B'$, along $B$ by $60^o$ to $C'$, along $C$ by $60^o$ to $A'$
We note that, 
$${\triangle APB} \cong ${\triangle ACB'}$$
$${\triangle APC} \cong ${\triangle A'CB}$$
$${\triangle BPC} \cong ${\triangle AC'B}$$
Thus we have, the area of the hexagon $AB'CA'BC'$
$$=2.{\triangle ABC}$$
We also observe that
${\triangle APB'}$, ${\triangle BPC'}$ and ${\triangle A'PC}$ are equilateral triangles
${\triangle PCB'} \cong {\triangle AC'P} \cong {\triangle A'BP} = RightAngle {\triangle}$
Thus we can say
$$2.{\triangle ABC} = [AB'CA'BC']$$
$$= {\triangle APB'} + {\triangle BPC'} + {\triangle A'PC} + 3.{\triangle PCB'}$$
$$= 12^2.\frac{\sqrt3}{4} + 13^2.\frac{\sqrt3}{4} + 5^2.\frac{\sqrt3}{4}  + 3.\frac{1}{2}(5)(12)$$
$$=169\frac{\sqrt3}{2} + 90$$
$${\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
Now, given the sides of the triangle $\overline {AB}$ = $\overline {BC}$ = $\overline {CA}$ = $a$, then
$$\frac{\sqrt3}{4}a^2 = {\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
$$\frac{\sqrt3}{4}a^2 =  169\frac{\sqrt3}{4} + 45$$
$$a^2 =  169 + 45\frac{4}{\sqrt3}$$
$$a = \sqrt{169 + 45\frac{4}{\sqrt3}}$$
Simplifying
$$a = \sqrt{169 + 60\sqrt3}$$
