If a secant and a tangent of a circle intersect in a point outside the circle then the area of the rectangle formed by the two line segments corresponding to the secant is equal to the area of the square formed by line segment corresponding the tangent.

If a secant and a tangent of a circle intersect in  a point outside the circle then the area of the rectangle formed by the two line segments corresponding to the secant is equal to the area of the square formed by line segment corresponding the tangent.

Given. PT is the tangent and PAB is a secant.

To Prove: PT² = PA × PB

Construction: Draw seg BT and AT

Solution:

In ∆ PTA and ∆ PBT

∠ PTA ≅ ∠ PBT [Angles in alternate segment]

∠ TPA ≅ ∠ TPB [Common angle]

∴ ∆PTA ∼ ∆ PBT  [A – A test for similarity]

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

∴ PT = PA PB PT

∴ PT2 = PA × PB

Hence Proved.