We know that,
[tex]EF=\dfrac{AD+BC}{2}[/tex]
which is
[tex]x=\dfrac{x-5+2x-4}{2}[/tex]
Now solve for x,
[tex]x=\dfrac{3x-9}{2}[/tex]
[tex]2x=3x-9[/tex]
[tex]x=9[/tex]
Since x is 9, the lengths are,
[tex]AD=x-5=9-5=\boxed{4}[/tex]
[tex]EF=x=\boxed{9}[/tex]
[tex]BC=2x-4=18-4=\boxed{14}[/tex]
Hope this helps :)
