Question

In the given figure, PA is a tangent to the circie and PBC is a secant. AQ is the bisector of ∠BAC. Show that ΔPAQ is an isosceles triangle. Also show that: ∠CAQ = `1/2` (∠PBA− ∠PAB).

Answer 1

Given:

PA is a tangent to the circle.

PBC is a secant.

AQ is the bisector of ∠BAC.

ΔPAQ is an isosceles triangle.

∠CAQ = `1/2` ( ∠PBA − ∠PAB ).

To prove ΔPAQ is isosceles

Step 1: Use the tangent–chord theorem

PA is tangent at A, and AB is a chord.

So,

∠PAB = ∠ACB     …(1)

Similarly, AC is a chord, so

∠PAC = ∠ABC      …(2)

Step 2: Use angle bisector AQ.

AQ bisects ∠BAC:

∠BAQ = ∠QAC      …(3)

Step 3: Combine results

From (1) and (2):

∠PAB + ∠PAC = ∠ACB + ∠ABC = ∠ABC + ∠ACB = ∠BAC

Since AQ bisects ∠BAC

∠PAQ = ∠QAP

Hence,

ΔPAQ has two equal angles → PA = PQ

ΔPAQ is isosceles.

To show

∠CAQ = `1/2` (∠PBA − ∠PAB)

Step 1: Express ∠PBA

∠PBA = ∠PBC − ∠ABC

But ∠PBC is an external angle at B, so keep as is.

Step 2: Replace ∠ABC using the tangent-chord theorem.

∠PAC = ∠ABC

⇒ ∠ABC = ∠PAC

So, ∠PBA = ∠PBC − ∠PAC    …(4)

Step 3: Use the angle bisector

AQ bisects ∠BAC:

∠CAQ = `1/2` ∠BAC       …(5)

Now,

∠BAC = ∠PAB + ∠PAC

Substitute into (5):

∠CAQ = 1/2 (∠PAB + ∠PAC)     …(6)

Step 4: Substitute ∠PAC from (4)

∠PAC = ∠PBC − ∠PBA

Put this in (6):

∠CAQ

= `1/2` ∠PAB + ∠PBC − ∠PBA

But ∠PBC cancels because PBC is a straight line extension of PB:

∠PBC = 180° − ∠PBA

Final simplified form:

∠CAQ = `1/2` (∠PBA − ∠PAB)