Find the length of the line

Since $\triangle ACD \sim \triangle AEF$ , we have: $\dfrac{AF}{EF}=\dfrac{AD}{DC}$ $\implies$ $AF=\dfrac{EF(AD)}{20}$ $\color{#D61F06}\boxed{1}$

Since $\triangle BAD \sim \triangle EFD$ , we have: $\dfrac{ED}{EF}=\dfrac{AD}{AB}$ $\implies$ $FD=\dfrac{EF(AD)}{15}$ $\implies$ $AD-AF=\dfrac{EF(AD)}{15}$ $\color{#D61F06}\boxed{2}$

Substitute $\color{#D61F06}\boxed{1}$ in $\color{#D61F06}\boxed{2}$ , we have

$AD-\dfrac{EF(AD)}{20}=\dfrac{EF(AD)}{15}$

Multiply both sides by the LCM of the denominators which is $60$ .

$\left[AD-\dfrac{EF(AD)}{20}\right]60=\left[\dfrac{EF(AD)}{15}\right]60$

$60AD-3(EF)(AD)=4(EF)(AD)$

$60AD=7(EF)(AD)$

$EF=\dfrac{60}{7}$

The desired answer is $(2a)^2+2b^2=(2\cdot 60)+2(7^2)=14547$

Note:

I multiplied both sides of the equation by the LCM of the denominators which is $60$ to remove the denominators (for easy calculation).

or you can do this

$AD-\dfrac{EF(AD)}{20}=\dfrac{EF(AD)}{15}$

$AD=\dfrac{EF(AD)}{15}+\dfrac{EF(AD)}{20}$

$AD=\dfrac{4(EF)(AD)+3(EF)(AD)}{60}$

$60AD=7(EF)(AD)$

$EF=\dfrac{60}{7}$

Which is the same, you must find the LCD so that the two fractions can be added.