# If PAB is a secant to a circle and PT is a tangent to the same circle then prove that:(PT)square=(PA)*(PB).

$\\ \text{Given : PAB is secant intersecting the circle with centre O at A} and B and a tangent PT at T. \\ Draw OM perpendicular to \mathrm{AB} and join \mathrm{OA}, \mathrm{OB} and \mathrm{OP} since, OM is perpendicular to \mathrm{AB} and a line drawn through the center is perpendicular to the chord, it bisects it also.\\ Therefore, \mathrm{AM}=\mathrm{MB}. \\ Now, \mathrm{PA} \times \mathrm{PB}=(\mathrm{PM}-\mathrm{AM}) \times (\mathrm{PM}+\mathrm{MB}) \\ since \mathrm{AM}=\mathrm{MB} \\ Therefore, \mathrm{PA} \times \mathrm{PB}=(\mathrm{PM}-\mathrm{AM}) \times (\mathrm{PM}+\mathrm{AM}) \\ \mathrm{PA} \times \mathrm{PB}=\mathrm{PM}^{2}-\mathrm{AM}^{2}-----------(1) \\ Also, since triangle (PMO) is a right angled triangle, therefore, by pythagoras theorem, \\ \mathrm{PM}^{2}=\mathrm{OP}^{2}-\mathrm{OM}^{2} \\ Also, triangle AMO is right angles triangle, therefore, \\ \mathrm{AM}^{2}=\mathrm{OA}^{2}-\mathrm{OM}^{2}$

$\\ \text{Substituting it back in ( 1 )}\\ \mathrm{PA} \times \mathrm{PB}=\mathrm{OP}^{2}-\mathrm{OM}^{2}-\left(\mathrm{OA}^{2}-\mathrm{OM}^{2}\right) \\ \mathrm{PA} \times \mathrm{PB}=\mathrm{OP}^{2}-\mathrm{OA}^{2} \\ \mathrm{PA} \times \mathrm{PB}=\mathrm{OP}^{2}-\mathrm{OT}^{2}\ (\mathrm{OA}=\mathrm{OT} as it is the radius of the circle) \\ \mathrm{PA} \times \mathrm{PB}=\mathrm{PT}^{2} (By using pythagoras theorm in triangle POT)$

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