A secant through point P intersects the circle in points A and B. Tangent drawn through P touches the circle at point T. Prove that PA × PB = PT^2

Given: A secant through point P intersects the circle in points A and B.

 Tangent drawn through P touches the circle at point T. Prove that PA × PB = PT²

Solution:

Given: Line PAB is a secant and Line PT is the tangent

To Prove: PA × PB = PT2

Construction: Draw chord BT and AT

Now, In ∆ PTA and ∆ BPT,

∠ PTA = ∠ PBT [Angles in alternate segment]

∠ APT = ∠ BPT [Common angle]

∴ ∆ PTA ∼ ∆ PBT

PT = TA = PA [C.S.S.T.] BP BT PT

PT = PA [C.S.S.T.] BP PT

∴ PT2 = PA × PB

Hence Proved