Question

In the given figure, $AB$ and $CD$ are two chords of a circle, intersecting each other at a point $E.$ Prove that $\angle AEC=\frac{1}{2}$(angle subtended by arc $CXA$ at the centre $+$ angle subtended by arc $DYB$ at the centre).

Answer 1

Construction: Join $AD$

To prove: $\angle AEC=\frac{1}{2}$(angle subtended by arc $CXA$ at the centre $+$ angle subtended by arc $DYB$ at the centre).

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc $AXC$ subtends $\angle AOC$ at the centre and $\angle ADC$ at $D$ on the circle.

$\therefore \angle AOC=2\angle ADC$

$\Rightarrow \angle ADC=\frac{1}{2}\angle AOC...\left(1\right)$

Also, arc DYB subtends \angle DOB at the centre and\angle DAB at A on the circle.

$\therefore \angle DOB=2\angle DAB$

$\Rightarrow \angle DAB=\frac{1}{2}\angle DOB...\left(2\right)$

Now, in $△ADE,$ we have

$\angle AEC=\angle ADC+\angle DAB$ (Exterior angle)

$\Rightarrow \angle AEC=\frac{1}{2}\angle AOC+\angle DOB$ [$\because \angle DAB=\angle DOB,$ angle subtended by same arc at different points on the circle are equal]

from $eq.\left(1\right)$ and $\left(2\right)$, we have

Hence, $\angle AEC=\frac{1}{2}$(angle subtended by arc $CXA$ at the centre + angle subtended by arc $DYB$ at the centre).