In the figure; PA is a tangent to the circle, PBC is secant and AD bisects angle BAC. Show that triangle PAD is an isosceles triangle. Also, show that :

∠CAD = $\dfrac{1}{2}$ [∠PBA - ∠PAB]

As, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :

From figure,

PA is a tangent and AB is a chord.

∴ ∠PAB = ∠C [Angles in alternate segment are equal]………..(1)

Given,

AD is bisector of ∠BAC.

∴ ∠BAD = ∠DAC …….(2)

We know that,

An exterior angle is equal to sum of two opposite interior angles.

⇒ ∠ADP = ∠C + ∠DAC

⇒ ∠ADP = ∠PAB + ∠BAD [From (1) and (2)]

⇒ ∠ADP = ∠PAD

Since, sides opposite to equal sides are equal.

∴ PA = PD

∴ PAD is an isosceles triangle.

In △ABC,

⇒ ∠PBA = ∠C + ∠BAC [Exterior angle is equal to sum of two opposite interior angles]

⇒ ∠BAC = ∠PBA - ∠C

⇒ ∠BAC = ∠PBA - ∠PAB [As, ∠C = ∠PAB]

⇒ 2∠CAD = ∠PBA - ∠PAB [As, AD bisects ∠BAC]

⇒ ∠CAD = $\dfrac{1}{2}$(∠PBA - ∠PAB).

Hence, proved that ∠CAD = $\dfrac{1}{2}$(∠PBA - ∠PAB).