In the given figure PT is a tangent to the circle. Chord BA produced meets the tangent PT at P. Given PT = 20 cm and PA = 16 cm.

(a) In △ PTB and △ PAT,

⇒ ∠PTA = ∠PBT (Alternate segment theorem)

⇒ ∠TPA = ∠BPT (Common angle)

∴ △ PTB ~ △ PAT (By A.A. axiom)

Hence, proved that △ PTB ~ △ PAT.

(b) We know that,

If a chord and a tangent intersect externally, then the product of the lengths of segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

⇒ PA × PB = PT²

⇒ PA × (PA + AB) = PT²

⇒ 16 × (16 + AB) = 20²

⇒ 16 × (16 + AB) = 400

⇒ 16 + AB = 25

⇒ AB = 25 - 16 = 9 cm.

Hence, AB = 9 cm.