In the figure (1) given below, AB, EF and CD are parallel lines. Given that AB = 15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate
(i) EF (ii) AC

(i) Consider △EFG and △CGD

∠ EGF = ∠ CGD [Vertically opposite angles]

∠ FEG = ∠ GCD [Alternate angles are equal]

So, by AA rule of similarity △EFG ~ △CGD.

Then,

$\dfrac{EG}{GC} = \dfrac{EF}{CD} \\[1em] \dfrac{5}{10} = \dfrac{EF}{18} \\[1em] EF = \dfrac{90}{10} \\[1em] EF = 9.$

Hence, the length of EF = 9 cm.

(ii) Consider △ABC and △EFC

∠ C = ∠ C [Common angles]

∠ ABC = ∠ EFC [Alternate angles are equal]

So, by AA rule of similarity △ABC ~ △EFC

Then,

$\dfrac{AC}{EC} = \dfrac{AB}{EF} \\[1em] \dfrac{AC}{EG + GC} = \dfrac{15}{9} \\[1em] \dfrac{AC}{5 + 10} = \dfrac{15}{9} \\[1em] \dfrac{AC}{15} = \dfrac{15}{9} \\[1em] AC = \dfrac{225}{9} \\[1em] AC = 25.$

Hence, the length of AC = 25 cm.