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Theorem 2

Theorem 2. If PAB is a secant to a circle intersecting it at A and B and PT is a tangent then PA.PB = PT2.

Given: - PAB is secant intersecting the circle with centre O at A and B and a tangent PT at T.

To Prove: - PA.PB = PT2             

Construction: - is drawn OA, OP and OT are joined.

Proof: -     PA = PM – AM

                  PB = PM + MB

                        

Also

PM2 = OP2 - OM2 [Pythagoras theo.]

and AM2 = OA2 - OM2 [Pythagoras theo.]

PA.PB = PT2 [Pythagoras theo.]

 

 


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