Solution for the value of angle $A$ of a triangle In triangle $\triangle \; ABC$ , if $$2\frac{\cos A}{a} + \frac{\cos B}{b} + 2\frac{\cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$$ find angle $A$.
This is a quiz bee problem sent to me by my friend in FB.  He asked me if I can do a solution for it.  Well I tried several ways but I am out of idea now.  The answer is 90 degree but what  he asked, and I am also asking it now, is the solution for it.
Thank you.

Using $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$ etc., 
we get, $$\frac2a\frac{b^2+c^2-a^2}{2bc}+\frac1b\frac{a^2+c^2-b^2}{2ac}+\frac2c\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2}{abc}$$
or, $$ 2(b^2+c^2-a^2)+(a^2+c^2-b^2)+2(a^2+b^2-c^2)=2(a^2+b^2)$$
or $b^2+c^2=a^2$ as $abc\ne0$ $a,b,c$ being the sides of triangle. 
So, $\cos A=0\implies A=(2n+1)\frac\pi2 $ where $n$ is any integer.
As $0<A<\pi,A=\frac\pi2$
