In an equilateral $\triangle ABC$ : $ DB^2 + DC^2 + BC^2 = 100 $ I have a question that:
There is a point $D$ inside the equilateral triangle ABC. If
$$ DB^2 + DC^2 + BC^2 = 100  $$
and the area of $DBC$ is $ 5 \sqrt{3} $, find $AD^2$.

This is what I tried: $DB = x ,\: DC = y,\: BC = a$. Then $$ x^2 + y^2 + a^2 = 100 \tag{1}$$
$$ \sqrt{ 2(x^2 y^2 + a^2 x^2 + a^2 y^2 )- (a^4 + x^4 + y^4 ) } = 20 \sqrt{3} \tag{2} $$
I want to find $a$, $x$, and $y$, but I have only two equations. What can I do?
 A: Denote $x= AD$, $y= BD$, $z=CD$, $a=BC(=AB=AC)$ and $\alpha=\angle BDC$. So we have: $$(*)\ y^2+z^2+a^2=100,$$
by the area of $\triangle DBC$ condition we have:
$$(**)\ yz\sin\alpha=10\sqrt 3,$$
and by cosine rule on $\triangle DBC$ we have:
$$(***)\ a^2=y^2+z^2-2yz\cos\alpha.$$
Consider the rotation $\mathcal R_{C,60^\circ}$ mapping $B\mapsto A$, $A\mapsto M$ and $D\mapsto N$. Since $CD=CN$ and $\angle DCN=60^\circ$, $\triangle DCN$ is equilateral, and thus $DN=DC=z$ and $\angle DNC=60^\circ$. Since we mapped $BD$ do $AN$, we have $AN=BD=y$. Also we mapped $\triangle BDC$ to $\triangle ANC$, so $\angle ANC=\angle BDC=\alpha$, wherefrom $\angle AND=\angle ANC-\angle DNC=\alpha-60^\circ$. So, in $\triangle AND$ we have $AD=x$, $AN=y$, $DN=z$ and $\angle AND=\alpha-60^\circ$, so by cosine rule:
$$(****)\ x^2=y^2+z^2-2yz\cos(\alpha-60^\circ)= y^2+z^2-yz(\cos\alpha+\sqrt 3\sin\alpha).$$
Multiply $(****)$ by $2$ and substract $(***)$. We get:
$$2x^2-a^2= y^2+z^2-2\sqrt 3yz\sin\alpha.$$
By using $(*)$ and $(**)$ we get:
$$2x^2= a^2+y^2+z^2-2\sqrt 3\cdot 10\sqrt 3=100-60=40,$$
so $AD^2=x^2=20$.
A: 
Let $|AB|=|BC|=|CA|=a$,
$|DB|=m$, $|DC|=n$, $|AD|=q$,
$|OB|=\tfrac12a$,
$|OH|=x$, $|DH|=y$
We have two constraints 
\begin{align}
m^2+n^2+a^2&=100
\tag{1}\label{1}
,\\
S_{DBC}&=5\sqrt3
\tag{2}\label{2}
.
\end{align} 
From \eqref{1}, 
\begin{align}
(\tfrac12a+x)^2+y^2+
(\tfrac12a-x)^2+y^2+a^2=100
,\\
x^2+y^2 = 50-\tfrac34a^2
\tag{3}\label{3}
,
\end{align}
hence one condition is that the point $D$ must be on a circle
with the center in the middle of $BC$.
Next, the area of $\triangle DBC$ is constrained,
\begin{align}
S_{ABC}&=\tfrac12 a y=3\sqrt5
,
\end{align}
and we can express the side of the $a$ inn terms of $y$,
\begin{align}
a&=\tfrac{10\sqrt3}y
.
\end{align}
Given that and assuming $x$ positive, 
we have
\begin{align}
x&=\frac1y\cdot\sqrt{50 y^2-225-y^4}=
\frac1y\cdot\sqrt{(45-y^2)(y^2-5)}
\\
&=
\frac1y\cdot\sqrt{(y-\sqrt5)(3\sqrt5-y)(y+\sqrt5)(3\sqrt5+y)}
,\\
\text{thus }\quad
y&\in(\sqrt5,3\sqrt5)
.
\end{align}
Next, 
\begin{align}
|AD|^2&=q^2=|AE|^2+|AD|^2
\\
&=
(\tfrac{\sqrt3}2\,a-y)^2+x^2
\tag{4}\label{4}
.
\end{align}
After the substitution of 
expressions for $a$ and $x$
in terms of $y$,
everything cancels nicely,
and we have the answer
\begin{align}
|AD|^2&=20
.
\end{align}
