In the given figure, AB || EF || DC; AB = 67.5 cm, DC = 40.5 cm and AE = 52.5 cm.

(i) Name the three pairs of similar triangles.

(ii) Find the length of EC and EF.

(i) The three pairs of similar triangle are :

In △BEF and △BDC

⇒ ∠FBE = ∠CBD [Common angle]

⇒ ∠BFE = ∠BCD [Corresponding angles are equal]

∴ △BEF ~ △BDC [By AA]

In △CEF and △CAB

⇒ ∠FCE = ∠BCA [Common angle]

⇒ ∠CFE = ∠CBA [Corresponding angles are equal]

∴ △CEF ~ △CAB [By AA]

In △ABE and △CDE

⇒ ∠AEB = ∠CED [Vertically opposite angles are equal]

⇒ ∠BAE = ∠ECD [Alternate angles are equal]

∴ △ABE ~ △CDE [By AA]

(ii) Since, △ABE and △CDE are similar,

$\therefore \dfrac{AB}{CD} = \dfrac{AE}{CE} \\[1em] \Rightarrow \dfrac{67.5}{40.5} = \dfrac{52.5}{CE} \\[1em] \Rightarrow CE = \dfrac{40.5 \times 52.5}{67.5} \\[1em] \Rightarrow CE = 31.5 \text{ cm}.$

Since, △CEF and △CAB are similar,

$\therefore \dfrac{CE}{CA} = \dfrac{EF}{AB} \\[1em] \Rightarrow \dfrac{31.5}{CE + AE} = \dfrac{EF}{67.5} \\[1em] \Rightarrow \dfrac{31.5}{31.5 + 52.5} = \dfrac{EF}{67.5} \\[1em] \Rightarrow EF = \dfrac{31.5 \times 67.5}{84} \\[1em] \Rightarrow EF = \dfrac{2126.25}{84} \\[1em] \Rightarrow EF = 25\dfrac{5}{16}\text{ cm}.$

Hence, CE = 31.5 cm and EF = $25\dfrac{5}{16}$ cm.