Question

Question 8
In the figure, 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

Construction: Extend the line DO and BO at the points l and H on the circle. Also, join AC.
We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it in the remaining part of the circle.

$\therefore \angle 1=2\angle 6...\left(i\right)$
and $\angle 3=2\angle 7...\left(ii\right)$

In $\Delta AOC,OC=OA$
[both are the radius of circle]

$\angle OCA=\angle 4$
[angles opposite to equal sides are equal]

Also,
$\angle AOC+\angle OCA+\angle 4={180}^{\circ }$
[by angle sum property of triangle]
$\Rightarrow \angle AOC+\angle 4+\angle 4={180}^{\circ }$
$\Rightarrow \angle AOC={180}^{\circ }-2\angle 4...\left(iii\right)$

Now, in $\Delta AEC,\angle AEC+\angle ECA+\angle CAE={180}^{\circ }$ [by angle sum propertyof a triangle]
$\Rightarrow \angle AEC={180}^{\circ }-(\angle ECA+\angle CAE)$
$\Rightarrow \angle AEC={180}^{\circ }-\left[\right(\angle ECO+\angle OCA)+\angle CAO+\angle OAE]$
$={180}^{\circ }-(\angle 6+\angle 4+\angle 4+\angle 5)$
[in $\Delta OCD,\angle 6=\angle ECO$, because, angles opposite to equal sides are equal]
$={180}^{\circ }-(2\angle 4+\angle 5+\angle 6)$
$={180}^{\circ }-({180}^{\circ }-\angle AOC+\angle 7+\angle 6)$
[From Eq. (iii) and in $\Delta AOB,\angle 5=\angle 7$, as angles opposite to equal sides are equal]
$=\angle AOC-\frac{\angle 3}{2}-\frac{\angle 1}{2}$ [From Eqs. (i) and (ii)]
$=\angle AOC-\frac{\angle 1}{2}-\frac{\angle 2}{2}-\frac{\angle 3}{2}+\frac{\angle 2}{2}$ $\left[addingandsubtracting\frac{\angle 2}{2}\right]$
$=\angle AOC-\frac{1}{2}(\angle 1+\angle 2+\angle 3)+\frac{\angle 8}{2}$ [$\because \angle 2=\angle 8$ (vertically opposite angles)]
$=\angle AOC-\frac{\angle AOC}{2}+\frac{\angle DOB}{2}$
$\Rightarrow \angle AEC=\frac{1}{2}(\angle AOC+\angle DOB)$
$=\frac{1}{2}$
[Angle sutended by arc CXA at the centre + angle subtended by arc DYB at the centre.]