# Area of Triangle in ellipse

Full question: Prove that the area of the triangle formed by three points of an ellipse, whose eccentric angles are $\theta , \phi$ and $\psi$ , is

$$\frac{1}{2}\sin\frac{\phi-\psi}{2}\sin\frac{\psi-\theta}{2}\sin\frac{\theta-\phi}{2}.$$

My attempt: I found the equation of the line joining the points with eccentric angles $\theta$ and $\phi$. Then found the perpendicular distance H of the point with ecc. angle $\psi$ from this line. Then I found the length D of the interval joining the first two points.

So the area is 1/2 DH, however I am unable to simply this expression into the form required.

If the coordinates of the triangle are given by $(x_a,y_a)$, $(x_b,y_b)$ and $(x_c,y_c)$, then the area of the triangle is obtain as $$\left\vert\dfrac{x_a(y_b-y_c) + x_b(y_c-y_a) + x_c(y_a-y_b)}2 \right\vert$$ In your case, we obtain $$\dfrac{ab}2\left\vert \cos(t_a)(\sin(t_b)-\sin(t_c)) + \cos(t_b)(\sin(t_c)-\sin(t_a)) + \cos(t_c)(\sin(t_a)-\sin(t_b))\right\vert$$ Use trigonometric identities and simplify.