Question

Theorem: 

Point E is in the exterior of a circle. A secant through E intersects the circle at points A and B, and a tangent through E touches the circle at point T, then EA × EB = ET².

Answer 1

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