Question

We have proved the above theorem by drawing seg AC and seg DB. Can the theorem be proved by drawing seg AD and seg CB, instead of seg AC and seg DB?

Answer 1

Yes, the theorem can be proved by drawing seg AD and seg CB.

Given: P is the centre of circle, chords AB and CD intersect internally at point E. 

To prove: AE × EB = CE × ED 

Construction: Draw seg AD and seg CB.

Proof: 

In ∆CEB and ∆AED, 

∠CEB = ∠DEA [Vertically opposite angles] 

∠CBE = ∠ADE [Angles inscribed in the same arc] 

∴ ∆CEB ~ ∆AED [by AA test of similarity] 

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

∴ AE × EB = CE × ED