If in $\Delta ABC,\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ca}$  then $\angle A=$

$\frac{2\left(b^2+c^2-a^2\right)+a^2+c^2-b^2+2\left(a^2+b^2-c^2\right)}{2abc}=\frac{a^2+b^2}{abc}$

$2b^2+2c^2-2a^2+a^2+c^2-b^2+2a^2+2b^2-2c^2=2a^2+2b^2$

$c^2+b^2=a^2$

$\therefore \angle A=90°$