if a secant and tangent intersect in a point outside in a circle .Then the area of the rectangle formed by the two line segments. Corresponding to the secant is eqaul to the area of the square formed by the line segment corresponding to the tangent ..
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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.
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.
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