Prove area of a quadrilateral is $\frac14[4m^2n^2-(b^2+d^2-a^2-c^2)^2]^{\frac12}$ Someone asked me this question which I am really stuck at, any help is appreciated.

If $a,b,c,d$ are the sides of a  quadrilateral and $m,n$ are diagonals of the quadrilateral, then prove that area of the quadrilateral is $$\frac14[4m^2n^2-(b^2+d^2-a^2-c^2)^2]^{\frac12}$$

I do not recall anything like this from my school days. I need some hints to get started on this, I do not remember high school geometry formulas, may be one of them could help here.
It works fine for squares and recatngles, so I assume it is a valid formula.

For convex $\square ABCD$, we can proceed as follows:

Let the diagonals meet a $P$, which subdivides the diagonals into segments of length $m_1$, $m_2$, $n_2$, $n_2$; and let the diagonals make an angle of $\theta$ at $P$ as shown. Then we can write:
$$\begin{align}
|\square ABCD| &= |\triangle APB| + |\triangle BPC| + |\triangle CPD| + |\triangle DPA| \\[4pt]
&= \frac12 \sin\theta \;\left( m_1 n_1 + m_2 n_1 + m_2 n_2 + m_1 n_2 \right) = \frac12 (m_1 + m_2)(n_1+n_2) \sin\theta \\[4pt]
&= \frac12 m n \sin\theta \\[4pt]
\therefore 4\;|\square ABCD| &= 2 m n \sin\theta \tag{$\star$}
\end{align}$$
Applying the Law of Cosines in each triangle gives:
$$\begin{align}
a^2 &= m_1^2 + n_1^2 - 2 m_1 n_1 \cos\theta \qquad\qquad
b^2 = m_2^2 + n_1^2 + 2 m_2 n_1 \cos\theta \\
c^2 &= m_2^2 + n_2^2 - 2 m_2 n_2 \cos\theta \qquad\qquad
d^2 = m_1^2 + n_2^2 + 2 m_1 n_2 \cos\theta
\end{align}$$
so that 
$$a^2 - b^2 + c^2 - d^2 = -2 m n \cos \theta$$
Finally, recalling $(\star)$,
$$\begin{align}
4 m^2 n^2 = 4 m^2 n^2\left( \sin^2\theta + \cos^2 \theta \right) = 16\;|\square ABCD|^2 + \left( a^2 - b^2 + c^2 - d^2 \right)^2
\end{align}$$
which gives the desired formula. $\square$
