If a, b and c are the lengths of the sides of a triangle ABC and if $p_1$, $p_2$ and $p_3$ are the lengths of the perpendiculars drawn from the circumcenter onto the sides BC, CA and AB respectively, then show that
$$\dfrac{a}{p_1} + \dfrac{b}{p_2} + \dfrac{c}{p_3}= \dfrac{abc}{4p_1 p_2 p_3}$$
I have done some bashing but I am unable to arrive at the final result. This problem is of 10+2 level as well.
We know
$ a=2R\sin A, b=2R\sin B, c=2R\sin C $
and
$ p_1=R\cos A, p_2=R\cos B, p_3=R\cos C $
This can be seen on a diagram of a triangle and its circumcircle. From the right triangles and little angle chasing we get these.