In the given figure, BA ∥ DC. Show that ΔOAB ∼ ΔODC. If AB = 4 cm, CD = 3 cm, OC = 5.7 cm and OD = 3.6 cm, find OA and OB.

Considering ΔAOB and ΔDOC,

∠AOB = ∠COD [Vertically opposite angles are equal]

∠A = ∠D [Alternate angles are equal]

∴ ΔAOB ∼ ΔDOC (By A.A. axiom)

We know that,

Corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{OA}{OD} = \dfrac{OB}{OC} = \dfrac{AB}{DC}$

Considering,

$\Rightarrow \dfrac{AB}{DC} = \dfrac{OA}{OD} \\[1em] \Rightarrow \dfrac{4}{3} = \dfrac{OA}{3.6} \\[1em] \Rightarrow OA = \dfrac{4 \times 3.6}{3} \\[1em] \Rightarrow OA = \dfrac{14.4}{3} \\[1em] \Rightarrow OA = 4.8 \text{ cm}.$

Considering,

$\Rightarrow \dfrac{AB}{DC} = \dfrac{OB}{OC} \\[1em] \Rightarrow \dfrac{4}{3} = \dfrac{OB}{5.7} \\[1em] \Rightarrow OB = \dfrac{4 \times 5.7}{3} \\[1em] \Rightarrow OB = \dfrac{22.8}{3} \\[1em] \Rightarrow OB = 7.6 \text{ cm}.$

Hence, OA = 4.8 cm and OB = 7.6 cm.