The correct option is C 25 cm
Given, AB∥EF∥CD
In △EFG and △CDG,
∠EFG=∠GDC (EF∥CD, alt. ∠s are equal)
∠EGF=∠CGD (vert. opp. ∠s)
∴ △EFG∼△GCD (By AA similarity)
∴ EGGC=EFDC⇒EF18=510⇒EF=9 cm
Now in △s ABC and EFC,
∠ACB=∠ECF (common)
∠ABC=∠EFC (AB∥EF, corr. ∠s are equal)
∴ △ABC∼△EFC (By AA similarity)
⇒ ACEC=ABEF⇒AC(EG+GC)=ABEF
⇒ AC(5+10)=159⇒AC=25 cm.