Given : PAB is secant intersecting the circle with centre O at A and B and a tangent PT at T.
Draw OM perpendicular to AB and join OA, OB and OP
Since, OM is perpendicular to AB and a line drawn through the center is perpendicular to the chord, it bisects it also. (Can be proven easily by RHS congruency)
Therefore, AM = MB.
Now, PA*PB = (PM-AM) * (PM+MB)
Since AM = MB
Therefore, PA*PB = (PM-AM) * (PM+AM)
PA*PB = PM2-AM2 (1)
Also, since triangle (PMO) is a right angled triangle, therefore, by pythagoras theorem,
PM2 = OP2 - OM2
Also, triangle AMO is right angles triangle, therefore, AM2 = OA2 - OM2
Substituing it back in (1)
PA*PB = OP2 - OM2 - (OA2 - OM2)
PA*PB = OP2 - OA2
PA*PB = OP2 - OT2 (OA = OT as it is the radius of the circle)
PA*PB = PT2 (By using pythagoras theorm in triangle POT)