Question

In a circle with center O, chords AB and CD intersect inside the circumference at E. Prove that ∠ AOC + ∠ BOD = 2∠ AEC.

Answer 1

AB and CD are two chords of a circle with centre O. They intersect at point E inside the circle.

Step 1: Use the Central Angle Theorem

Arc AC subtends

∠AOC at the centre

∠ABC at the circumference

∠AOC = 2∠ABC       ...[1]

Arc BD subtends

∠BOD at the centre

∠BCD at the circumference

∠BOD = 2∠BCD        ...[2]

Step 2: Add [1] and [2]

∠AOC + ∠BOD = 2(∠ABC + ∠BCD)     ...[3]

Step 3: Use intersecting chords theorem

For two chords, AB and CD, intersecting at E inside the circle:

∠AEC = ∠ABC + ∠BCD

∠AOC + ∠BOD = 2∠AEC   ...[Hence proved]