Question

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

Answer 1

In the given figure, the same arc AC subtends $\angle AOC$ on the centre and $\angle ABC$ on the other part.

$\Rightarrow \angle AOC=2\angle ABC$ [Angle subtended on the centre of the circle is double the angle subtended on the other part of the circle by same arc]

Similarly, arc BD subtends $\angle BOD$ and $\angle DCB$

$\Rightarrow \angle BOD=2\angle DCB$

On adding the above results, we get

$\angle AOC+\angle BOD=2\angle ABC+2\angle DCB$

$=2(\angle ABC+\angle DCB)......\left(1\right)$

In $△ECB$, by exterior angle property, we have

$\angle AEC=\angle ECB+\angle EBC$

$\Rightarrow \angle AEC=\angle DCB+\angle ABC......\left(2\right)$

On putting the value of (2) in (1), we get

$\angle AOC+\angle BOD=2\angle AEC$

Hence proved.