In the figure (2) given below, CA ∥ BD, the lines AB and CD meet at O.

(i) Prove that △ACO ~ △BDO.

(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.

Considering △ACO and △BDO,

∠ AOC = ∠ BOD [Vertically opposite angles]
∠ A = ∠ B [Alternate angles]

Then, by AA rule of similarity, △AOC ~ △BOD.

So,

$\Rightarrow \dfrac{OA}{OB} = \dfrac{OC}{OD} = \dfrac{AC}{BD} \\[1em] \Rightarrow \text{Consider, } \dfrac{AC}{BD} = \dfrac{OA}{OB} \\[1em] \Rightarrow \dfrac{3.6}{2.4} = \dfrac{OA}{3.2} \\[1em] OA = \dfrac{3.6 \times 3.2}{2.4} \\[1em] OA = \dfrac{11.52}{2.4} \\[1em] OA = 4.8.$

Now, consider

$\dfrac{AC}{BD} = \dfrac{OC}{OD} \\[1em] \dfrac{3.6}{2.4} = \dfrac{OC}{4} \\[1em] OC = \dfrac{14.4}{2.4} \\[1em] OC = 6.$

Hence, OA = 4.8 cm and OC = 6 cm.