Question

In the adjoining figure, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If ∠AEP ≅ ∠DEP, then prove that AB = CD.

Given: P is the centre of the circle.

Chord AB and chord CD intersect on the diameter at the point E. ∠AEP ≅ ∠DEP 

To prove: AB = CD 

Construction: Draw seg PM ⊥ chord AB, A-M-B seg PN ⊥ chord CD, C-N-D

Comments

Thanks for the answer.......

Answer 1

Proof:

∠AEP ≅ ∠DEP [Given] 

∴ Seg ES is the bisector of ∠AED.

PoInt P is on the bisector of ∠AED.

∴ PM = PN [Every point on the bisector of an angle is equidistant from the sides of the angle.] 

∴ chord AB ≅ chord CD [Chords which are equidistant from the centre are congruent.] 

∴ AB = CD [Length of congruent segments]