find AB in trapezoid

We note that $\triangle EDC$ , $\triangle EGF$ , and $\triangle EAB$ are similar. Then

$\begin{aligned} \frac {AB}{GF} & = \frac {AE}{EF} = \frac {AF-EF}{EF} = \frac \blue{AF}{EF} -1 & \small \blue{\text{Note that }AF=CF} \\ & = \frac \blue{CF}{EF} - 1 = \frac {EC-EF}{EF} -1 = \blue{\frac {EC}{EF}} - 2 & \small \blue{\text{and that }\frac {EC}{EF} = \frac {CD}{FG}} \\ & = \blue{\frac {CD}{FG}} - 2 \\ \implies AB & = FG \left(\frac {CD}{FG} - 2\right) = 3 \left(\frac {97}3 - 2\right) = \boxed{91} \end{aligned}$

It is well known formula for trapezoid. Distance between the mid-point of diagonals of trapezoid is half the difference of parallel sides.

$\large\frac{CD - AB}{2}$ $= GF$

Putting $GF = 3$ and $CD = 97$ we get $AB = 91$