an equilateral ΔABC

Extend $AF,BF,CF$ and intersect circle $(DEFG)$ again at $A',B',C'$ , respectively.

Take a homothety at F and it will take $\triangle ABC$ to $\triangle A'B'C'$ . Let the ratio of the radius of the two circles be $\alpha$ .

$c^2=AD^2=AF*FA'=AF*\alpha*AF$ , so $c=\sqrt{\alpha}AF$

Similarly, $b=\sqrt{\alpha}CF$ , $a=\sqrt{\alpha}BF$ .

So, $\frac{a+b}c=\frac{CF+BF}{AF}$ , but by ptolemy theorem we know that $CF*AB+BF*AC=AF*BC$ ,

which is equivalent to $CF+BF=AF$ . Hence the desired ratio is equaled to 1.