In the given diagram, PT is a tangent to the circle and QT is a secant meeting the circle at R, such that QR = 9 cm and RT = 16 cm.

(a) In ∆PTR and ∆QTP,

⇒ ∠PTR = ∠QTP [Common angle]

⇒ ∠TPR = ∠TQP [Angle between tangent and chord is equal to the angle in the alternate segment]

∴ ∆PTR ~ ∆QTP (By A.A. axiom)

Hence, proved that ∆PTR ~ ∆QTP.

(b) Since, ∆PTR ~ ∆QTP, their corresponding sides are proportional.

$$\begin{aligned}\Rightarrow \frac{PT}{QT} = \frac{TR}{TP} \\[1em] \Rightarrow PT^2 = QT \times TR.\end{aligned}$$

From figure,

QT = QR + RT = 9 + 16 = 25 cm.

Substituting the values,

$$\begin{aligned}\Rightarrow PT^2 = 25 \times 16 \\[1em] \Rightarrow PT^2 = 400 \\[1em] \Rightarrow PT = \sqrt{400} \\[1em] \Rightarrow PT = 20 \text{ cm}.\end{aligned}$$

Hence, the length of PT = 20 cm.