In the given figure, two chords AC and BD of a circle intersect at E. If arc AB = CD, prove that : BE = EC and AE = ED.

Join AB and CD.

Given, arc AB = arc CD.

Since, equal arcs subtend equal chords.

⇒ chord AB = chord CD

In △ AEB and △ DEC,

⇒ AB = CD (Proved above)

⇒ ∠BAE = ∠CDE (Angles in the same segment subtended by arc BC)

⇒ ∠ABE = ∠DCE (Angles in the same segment subtended by arc AD).

By ASA Congruence,

∴ △ AEB ≅ △ DEC

Since corresponding parts of congruent triangles are equal.

⇒ BE = EC and AE = ED.

Hence, proved that BE = EC and AE = ED.