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Theorem: 

If secants containing chords AB and CD of a circle intersect outside the circle in point E, then AE × EB = CE × ED.

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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

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