Solving for angle $A$ in triangle ABC

Given:
$\dfrac{2 \cos A}{a} + \dfrac{\cos B}{b} + \dfrac{2 \cos C}{c} = \dfrac{a}{bc} + \dfrac{b}{ca}$
 

From Cosine law:
$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$

$\cos B = \dfrac{a^2 + c^2 - b^2}{2ac}$

$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$
 

Substitute to the given equation:
$\dfrac{2}{a} \cdot \dfrac{b^2 + c^2 - a^2}{2bc} + \dfrac{1}{b} \cdot \dfrac{a^2 + c^2 - b^2}{2ac} + \dfrac{2}{c} \cdot \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{a}{bc} + \dfrac{b}{ca}$

$\dfrac{b^2 + c^2 - a^2}{abc} + \dfrac{a^2 + c^2 - b^2}{2abc} + \dfrac{a^2 + b^2 - c^2}{abc} = \dfrac{a^2 + b^2}{abc}$

$(b^2 + c^2 - a^2) + \dfrac{a^2 + c^2 - b^2}{2} + (a^2 + b^2 - c^2) = a^2 + b^2$

$b^2 - a^2 + \dfrac{a^2 + c^2 - b^2}{2} = 0$

$2b^2 - 2a^2 + (a^2 + c^2 - b^2) = 0$

$b^2 - a^2 + c^2 = 0$

$b^2 + c^2 = a^2$   ←   Equation (1)
 

Equation (1) satisfies the Pythagorean theorem for a right triangle whose perpendicular sides are $b$ and $c$ and hypotenuse $a$. Thus, angle $A = 90^\circ$.   ←   answer