Let ∠CAB be 2x and ∠BAP y
Here, AD bisects ∠CAB
So, ∠CAD = ∠DAB = 1/2 ∠CAB = x ….(i)
Given, PA is the tangent and BA is the secant, therefore,
∠BAP = ∠BCA = y (Alternate segment theorem) ….(ii)
Now, ∠ADP is the exterior angle of triangle CAD
So, ∠ADP = x + y …..(iii)
(i)In triangle PAD,
∠PAD = ∠DAB + ∠BAP = x + y ….[From i & ii]
∠PDA = x + y ….[From iii]
Therefore, triangle PAD is isosceles.
(ii)∠PBA is the exterior angle of triangle ABC
So, ∠PBA = 2x + y
Substituting the values of ∠PBA and ∠PAB in the RHS of the statement, we get
RHS = 1/2 (∠PBA – ∠PAB) = 1/2 (2x + y – y) = x = ∠CAD = LHA
Thus, ∠CAD = 1/2 (∠PBA – ∠PAB)

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