Given: Secant through point E intersects the circle in points A and B.
Tangent drawn through point E touches the circle in point T.
To prove: EA × EB = ET2
Construction: Draw seg TA and seg TB.
Proof: In ∆EAT and ∆ETB,
∠AET ≅ ∠TEB [Common angle]
∠ETA ≅ ∠EBT [Theorem of angle between tangent and secant, E – A – B]
∴ ∆EAT ~ ∆ETB [AA test of similarity]
∴ EA/ET = ET/EB [Corresponding sides of similar triangles]
∴ EA × EB = ET2