Answer:
[tex]EF=\dfrac{120}{7}\ ft[/tex]
Step-by-step explanation:
Triangles CEF and CAD are similar right triangles (they have the common angle, so by AA postulate they are similar). Similar triangles have proportional corresponding sides, so
[tex]\dfrac{EF}{AD}=\dfrac{CF}{CD}\\ \\\dfrac{EF}{40}=\dfrac{CF}{CD}[/tex]
Triangles DEF and DBC are similar right triangles (they have the common angle, so by AA postulate they are similar). Similar triangles have proportional corresponding sides, so
[tex]\dfrac{EF}{BC}=\dfrac{DF}{DC}\\ \\\dfrac{EF}{30}=\dfrac{CD-CF}{CD}\\ \\\dfrac{EF}{30}=1-\dfrac{CF}{CD}[/tex]
Substitute the fraction [tex]\frac{CF}{CD}[/tex] from the first equality into the second equality:
[tex]\dfrac{EF}{30}=1-\dfrac{EF}{40}\ [\text{Muliply by 120}]\\ \\4EF=120-3EF\\ \\4EF+3EF=120\\ \\7EF=120\\ \\EF=\dfrac{120}{7}=17\dfrac{1}{7}\ ft[/tex]
