**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.

**Proof:**

In ∆ADE and ∆CBE,

∠AED = ∠CEB [Common angle]

∠DAE ≅ ∠BCE [Angles inscribed in the same arc]

∴ ∆ADE ~ ∆CBE [AA test of similarity]

∴ AE/CE = ED/EB [Corresponding sides of similar triangles]

∴ AE × EB = CE × ED