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

Question

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

Answer 1

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 testof similaritv] 

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

∴ AE × EB = CE × ED