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 $\overline{AB}$.

If we are allowed to use Coordinate Geometry,

WLOG we can assume the coordinates of $A,B,C$ to be $(x,y),(-a,0),(a,0)$ respectively and $D$ to be $(h,k).$
As $\triangle  ABC$ is equilateral $(x-a)^2+(y-0)^2=(x+a)^2+(y-0)^2=(-a-a)^2+(0-0)^2$
From the first relation we find $4ax=0\implies x=0$ as $a\ne0$
From the second relation we find $y^2=3a^2\implies y=\sqrt3a$ assuming  $a>0$
So, $\overline{AB}=2a$
So, $\overline{AD}^2=(h-0)^2+(k-\sqrt3a)^2$ But $\overline{AD}=13$
$\implies (h-0)^2+(k-\sqrt3a)^2=13^2--->(1)$
Similarly, for $BD,CD$ respectively,
$(h+a)^2+k^2=12^2--->(2)$ and  $(h-a)^2+k^2=5^2--->(3)$
So here we have three equations with three unknowns $a,h,k$
One way to solve this could be  as follows :
From $(2)-(3),4ah=119\implies h=\frac{119}{4a}$
From $(2)-(1),2ah+2\sqrt3ak=2a^2-25\implies k=\frac{4a^2-169}{4\sqrt3a} $ (putting $ah=\frac{119}4$)
Putting the values of $h,k$ in terms of $a$ in $(1),$
$$\left(\frac{119}{4a}\right)^2+\left(\frac{4a^2-169}{4\sqrt3a}-\sqrt3a\right)^2=13^2$$
$$\implies 64a^4-32\cdot169a^2+169^2+3(119^2)=0$$
which is a Quadratic Equation in $a^2$ 
