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If PAB is a secant to a circle and PT is a tangent to the same circle then prove that:

(PT)square=(PA)*(PB).

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(PT)square=(PA)*(PB).

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Join point T to point A and also to point B.

Now you observe two triangles 1. triangle PTA 2. triangle PBT

Also remember:Angle between a chord and a tangent at the point of contact

= angle,subtended by the same chord, at any point in the opposite segment.

TA is secant. TP is a tangent.

Therefore angle PTA (angle between tangent and chord)

= angle TBA (angle in the opposite segment) ----------------(1)

angle PTA = angle TBP ------------------------------------(2)

angle PTA in triangle PTA = angle TBA in triangle PBT [from (1) above] ---(3)

angle TPA in triangle PTA = angle TPB in triangle PBT (common angle) --(4)

From (3) and (4), two angles of triangles PTA and PBT are equal,

hence triangles PTA and PBT are similar.

Therefore PT / PB = PA / PT OR PT² = PAxPB