Area of triangle inscribed in a circle with a specific position Triangle $ABC$ inscribed in a circle. Versine (green)are drawn from the midpoints of the sides of the triangle perpendicular to them. They have lengths as shown in this figure.

find the area of $\Delta ABC$
honestly, I got stuck on this problem and I was far from geometry for years. Please show me a clue or guide me to get over on this problem. Thanks in advance.
I just find out the lines(green) must cross at one point, because of the middle and perpendicularly. but not go more ...
 A: Let $\alpha, \beta, \gamma = 16,20,13$ be the heights of circular segment opposite to vertices $A,B,C$. 
Let $R$ be the circumradius. It is easy to see
$$\alpha = R(1-\cos A),\quad \beta = R(1-\cos B) \quad\text{ and }\quad \gamma = R(1-\cos C)$$
Notice for any $3$ angles $A,B,C$ which sums to $180^\circ$, we have the trigonometry identity:
$$1 - \cos^2A - \cos^2 B - \cos^2 C - 2\cos A \cos B \cos C = 0\\
\iff 2(1-\cos A)(1-\cos B)(1-\cos C) = (1- \cos A - \cos B - \cos C)^2$$
In terms of $\alpha,\beta,\gamma$, this leads to
$$2\frac{\alpha\beta\gamma}{R^3} = \left(\frac{\alpha+\beta+\gamma}{R} - 2\right)^2
\quad\iff\quad (2R - (\alpha+\beta+\gamma))^2 R - 2\alpha\beta\gamma = 0
$$
Plug back the values of $\alpha,\beta,\gamma$, this becomes
$$R(2R - 49)^2 - 8320 = (2R-65)(2R^2 - 33R + 128) = 0$$
This cubic equations has $3$ real roots. However, the two roots from the quadratic factor is too small (both $\le 20$). This leaves us with $R = \frac{65}{2}$.
Apply intersecting chord theorem to side $BC$ and its perpendicular bisector, we find
$$\left(\frac{a}{2}\right)^2 = \alpha(2R - \alpha) \implies
a = 2\sqrt{\alpha(2R-\alpha)} = 2\sqrt{16(65-16)} = 56$$
By a similar argument, we find
$$b = 2\sqrt{20(65-20)} = 60\quad\text{ and }\quad c = 2\sqrt{13(65-13)} = 52$$
By Euler's formula between a triangle's area, circumradius and sides, the desired area equals to
$$ \verb/Area/(ABC) = \frac{abc}{4R} = \frac{56\cdot 60 \cdot 52}{2\cdot 65} = 1344$$
A: Hint: One can easily check that $R=\dfrac{65}2$ is the only solution of the equation
$$
\arccos\left(1-\frac{13}R\right)+\arccos\left(1-\frac{16}R\right)+\arccos\left(1-\frac{20}R\right)=\pi.
$$
