Question

In the given 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.

Answer 1

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 such that A-M-B and draw seg PN ⊥ chord CD such that C-N-D

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 equidistant from the center of the circle are congruent.)

∴ AB = CD      ...(Length of congruent segments)