Question

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

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Answer 1

Consider arc AC of the circle with centre at O. Clearly, arc AC subtends $\angle AOC$ at the centre and $\angle ABC$ at the remaining part of the circle.

$\angle AOC=2\angle ABC$ ......(i) [angle subtended at the centre is double than the angle subtended at the remaining part of circle]

Similarly, arc BD subtends $\angle BOD$ at the centre and $\angle BCD$ at the remaining part of the circle.

$\therefore \angle BOD=2\angle BCD$ ......(ii)[angle subtended at the centre is double than the angle subtended at the remaining part of circle]

Adding (i) and (ii), we get

$\angle AOC+\angle BOD=2(\angle ABC+\angle BCD)$

$\angle AOC+\angle BOD=2\angle AEC$ [Exterior angle property]

$(\because \angle ABC+\angle BCD=\angle AEC)$
Hence proved.