How to solve the following geometry problem?

Question

The diameters of the circumcircle of triangle $ABC$ drawn from $A$,$B$ and $C$ meet $BC$, $CA$ and $AB$, respectively, at $L$, $M$ and $N$. Prove that: $$\frac{1}{ AL} + \frac{1}{BM}+\frac{1}{CN}=\frac{2}{R}$$ I tried with formulas based on properties of triangle and some with geometry properties with circumcircle but at the end not able to reach the result, I earnestly urge for your help.

Answer 1

We have $$\dfrac{OA}{AL}=\dfrac{[AOC]}{[ALC]}=\dfrac{[AOB]}{[ALB]}$$$$=\dfrac{[AOC]+[AOB]}{[ALC]+[ALB]}=\dfrac{[AOC]+[AOB]}{[ABC]}$$

Similarly $$\dfrac{OB}{BM}=\dfrac{[AOB]+[BOC]}{[ABC]}$$

$$\dfrac{OC}{CN}=\dfrac{[BOC]+[AOC]}{[ABC]}$$

Summing up we get $$\dfrac{R}{AL}+\dfrac{R}{BM}+\dfrac{R}{CN}=\dfrac{2[ABC]}{[ABC]}=2$$