Given: Chords AB and CD of a circle intersect outside the circle in point E.
To prove: AE × EB = CE × ED
Construction: Draw seg AD and seg BC.
In ∆ADE and ∆CBE, ∠AED = ∠CEB [Common angle] ∠DAE ≅ ∠BCE [Angles inscribed in the same arc]
∴ ∆ADE ~ ∆CBE [AA testof similaritv]
∴ AE/CE = ED/EB [Corresponding sides of similar triangles]
∴ AE × EB = CE × ED