Flag problem

BY british flag theorem,
$ae^2$ + $ce^2$ = $be^2$ + $de^2$
Therefore, 53+52=85+ $be^2$
be= $\sqrt{20}$

Draw two parallel lines as above. Here we will see that $r^2+g^2 = 53; r^2+y^2 = 85; y^2+b^2 = 52$ As we need to find $\sqrt{g^2+b^2}$ , we just combine all variables we know and eliminate what we don't need, which gives $\sqrt{(r^2+g^2)+(y^2+b^2)-(r^2+y^2)} = \sqrt{53+52-85} = \sqrt{20}$