In the given diagram O is the centre of the circle. Chord SR produced meets the tangent XTP at P.

(a) We know that,

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment of the circle.

∴ ∠PTR = ∠PST

⇒ ∠RPT = ∠TPS [Common angles]

Therefore, by AA similarity, ΔPTR ~ ΔPST.

Hence, proved that ΔPTR ~ ΔPST.

(b) Since, corresponding sides of similar triangles are proportional we have :

⇒ $\frac{PT}{PS} = \frac{PR}{PT}$

⇒ PT² = PR × PS.

Hence, proved that PT² = PR × PS.

(c) Given,

PR = 4 cm and PS = 16 cm.

$$\begin{aligned}\Rightarrow PT^2 = PR \times PS \\[1em] \Rightarrow PT = \sqrt{PR \times PS} \\[1em] \Rightarrow PT = \sqrt{4 \times 16} \\[1em] \Rightarrow PT = \sqrt{64} \\[1em] \Rightarrow PT = 8 \text{ cm}.\end{aligned}$$

Hence, PT = 8 cm.