show that the points P,B, Q and T are concyclic

Question

two circles intersect each other at points A and B. Astraight line PAQ cuts the circles at P and Q. if the tangents P and Q intersects at point T. prove the above

Answer 1

Problem restated

1. Two circles intersect at points A and B.
2. A straight line PAQ cuts the circles at P (on the first circle) and Q (on the second circle).
3. Tangents at P and Q meet at T.

We need to prove: P, B, Q, T are concyclic (lie on a circle).

Step 1: Recall tangent property

If a tangent touches a circle at point P, then the angle between the tangent and a chord through P equals the angle in the opposite arc.

That is, by the alternate segment theorem:

∠(TP, chord PA)=∠PBA

(because PA is a chord and T is tangent at P).

So:

∠APT=∠PBA

Similarly, tangent at Q:

∠AQT=∠QBA

Step 2: Add the two results

From (1) and (2):

1. ∠APT = ∠PBA
2. ∠AQT = ∠QBA

So:

∠APT+∠AQT=∠PBA+∠QBA

But notice:

1. ∠APT + ∠AQT = ∠PTQ (since P–A–Q is a straight line).
2. ∠PBA + ∠QBA = ∠PBQ.

So:

∠PTQ=∠PBQ

Step 3: Concyclic condition

If four points P, B, Q, T are concyclic, then the opposite angles at T and B are equal (or supplementary).

Here we have shown:

∠PTQ=∠PBQ

That’s exactly the angle condition for cyclic quadrilaterals.

Therefore, P, B, Q, T are concyclic.