In the adjoining figure, chords AB and CD of the circle are produced to meet at O. Prove that triangles ODB and OAC are similar. Given that CD = 2 cm, DO = 6 cm and BO = 3 cm, calculate AB. Also find $\dfrac{\text{Area of quad. CABD}}{\text{Area of △OAC}}.$

In △ODB and △OAC,

∠ODB = ∠C

∠O = ∠O (Common)

∴ △ODB ~ △OAC (AA axiom)

Since, in similar triangles the ratio of the corresponding sides are equal.

$\therefore \dfrac{OD}{OA} = \dfrac{OB}{OC} \\[1em] \Rightarrow \dfrac{6}{OA} = \dfrac{3}{6 + 2} \\[1em] \Rightarrow OA = \dfrac{6 \times 8}{3} \\[1em] \Rightarrow OA = 16.$

AB = OA - OB = 16 - 3 = 13 cm.

Since, △ODB ~ △OAC

$\therefore \dfrac{\text{Area of △OAC}}{\text{Area of △ODB}} = \dfrac{OC^2}{OB^2} \\[1em] \dfrac{\text{Area of △OAC}}{\text{Area of △ODB}} = \dfrac{8^2}{3^2} \\[1em] \dfrac{\text{Area of △OAC}}{\text{Area of △ODB}} = \dfrac{64}{9} …(i)$

Subtracting 1 from both sides we get,

$\dfrac{\text{Area of △OAC}}{\text{Area of △ODB}} - 1 = \dfrac{64}{9} - 1 \\[1em] \dfrac{\text{Area of △OAC - Area of △ODB}}{\text{Area of △ODB}} = \dfrac{64 - 9}{9} \\[1em] \dfrac{\text{Area of quadrilateral CABD}}{\text{Area of △ODB}} = \dfrac{55}{9} ….(ii)$

Dividing (ii) by (i),

$\dfrac{\dfrac{\text{Area of quadrilateral CABD}}{\text{Area of △ODB}}}{\dfrac{\text{Area of △OAC}}{\text{Area of △ODB}}} = \dfrac{\dfrac{55}{9}}{\dfrac{64}{9}} \\[1em] \dfrac{\text{\text{Area of quadrilateral CABD}}}{\text{Area of △OAC}} = \dfrac{55}{64}.$

Hence, the length of AB = 13 cm and

$\dfrac{\text{\text{Area of quadrilateral CABD}}}{\text{Area of △OAC}} = \dfrac{55}{64}.$