Question

In Fig. 10.19 AB and CD are two chords of a circle intersecting each other at point E Prove that $\angle AEC=\frac{1}{2}$(Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre)

Answer 1

$AB$ and $CD$ are two chords of a circle intersecting each other at point $E$.

We have to prove that $\angle AEC=\frac{1}{2}$ (Angles subtended by an arc $CXA$ at the centre + angle subtended by arc $DYB$ at the centre.)

Join $AC.BC$ and $BD$

Since, the angles subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. now arc $CXA$ subtends $\angle AOC$ at the centre and $\angle ABC$ at the remaining pan of the circle, so

$\angle AOC=2\angle ABC$.........(1)

Similarly, $\angle BOD=2\angle BCD$...........(2)

Now, adding (1) and (2), we get

$\angle AOC+\angle BOD=2(\angle ABC+\angle BCD)$........(3)

Since exterior angle of a triangle is equal to the sum of interior opposite angles,

so in $\Delta CEB$, we have

$\angle AEC+\angle ABC+\angle BCD$..........(4)

From (3) and (4), we get

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

or $\angle AEC=\frac{1}{2}(\angle AOC+\angle BOD)$

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