# In the fig., PA is a tangent to the circle PBC is secant and AD bisects angle BAC. Show that triangle PAD is an isosceles triangle.

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)