Question
In Fig. $AB$ and $CD$ are two chords of a circle intersecting each other at point $E.$ Prove that $\angle\text{AEC}=\frac{1}{2}$ $($Angle subtended by arc $CXA$ at centre $+$ angle subtended by arc $DYB$ at the centre$).$
✓
Answer
Given: In a figure, two chords $AB$ and $CD$ intersecting each other at point $E.$
To prove: $\angle\text{AEC}=\frac{1}{2} [$angle subtended by arc $C \times A$ at centre $+$ angle subtended by arc $DYB$ at the centre$]$
Construction: Extend the line $DO$ and $BO$ at the points $l$ and $H$ on the circle. Also, join $AC.$
Proof: We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\therefore\angle1=2\angle6\ \ ...(\text{i})$ and $\angle3=2\angle7\ \ ...(\text{ii)}$ In $\triangle\text{AOC},\ \ \text{OC}=\text{OA}$ $[$both are the radius of circle$]$
$\angle\text{OCA}=\angle4$ [angles opposite to equal sides are equal] Also,
$\angle\text{AOC}+\angle\text{OCA}+\angle4=180^\circ$ $[$by angle sum property of triangle$]$
$\Rightarrow\angle\text{AOC}+\angle4+\angle4=180^\circ$
$\Rightarrow\angle\text{AOC}=180^\circ-2\angle4\ \ \ ...(\text{iii})$
Now, in $\triangle\text{AEC},\ \ \ \angle\text{AEC}+\angle\text{ECA}+\angle\text{CAE}=180^\circ$ [by angle property sum of a triangle]
$\Rightarrow\angle\text{AEC}=180^\circ-(\angle\text{ECA}+\angle\text{CAE})$
$\Rightarrow\angle\text{AEC}=180^\circ-[(\angle\text{ECO}+\angle\text{OCA})+\angle\text{CAO}+\angle\text{OAE}]$
$=180^\circ-(\angle6+\angle4+\angle4+\angle5)$
$\big[$In $\triangle\text{OCD},\angle6=\angle\text{ECO}$ angles opposite to equal sides are equal$\big]$
$=180^\circ-(2\angle4+\angle5+\angle6)$
$=180^\circ-(180^\circ-\angle\text{AOC}+\angle7+\angle6)[ $From Eq. $(iii)$ and in
$\triangle\text{AOB}.\angle5=\angle7,$ as $($angles opposite to equal sides are equal$)]$
$=\angle\text{AOC}-\frac{\angle3}{2}-\frac{\angle1}{2} [$from Eqs. $(i)$ and $(ii)]$
$=\angle\text{AOC}-\frac{\angle1}{2}-\frac{\angle2}{2}-\frac{\angle3}{2}+\frac{\angle2}{2}$ [adding and subtracting $\frac{\angle2}{2}$]
$=\angle\text{AOC}-\frac{1}{2}(\angle1+\angle2+\angle3)+\frac{\angle8}{2} [ \because\angle2=\angle8$ (vertically opposite angles)]
$=\angle\text{AOC}=\frac{\angle\text{AOC}}{2}+\frac{\angle\text{DOB}}{2}$
$\Rightarrow\angle\text{AEC}=\frac{1}{2}(\angle\text{AOC}+\angle\text{DOB})$
$=\frac{1}{2} [$angle subtended by arc $CXA$ at the centre $+$ angle subtended by arc $DYB$ at the centre$]$
To prove: $\angle\text{AEC}=\frac{1}{2} [$angle subtended by arc $C \times A$ at centre $+$ angle subtended by arc $DYB$ at the centre$]$
Construction: Extend the line $DO$ and $BO$ at the points $l$ and $H$ on the circle. Also, join $AC.$
Proof: We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\therefore\angle1=2\angle6\ \ ...(\text{i})$ and $\angle3=2\angle7\ \ ...(\text{ii)}$ In $\triangle\text{AOC},\ \ \text{OC}=\text{OA}$ $[$both are the radius of circle$]$
$\angle\text{OCA}=\angle4$ [angles opposite to equal sides are equal] Also,
$\angle\text{AOC}+\angle\text{OCA}+\angle4=180^\circ$ $[$by angle sum property of triangle$]$
$\Rightarrow\angle\text{AOC}+\angle4+\angle4=180^\circ$
$\Rightarrow\angle\text{AOC}=180^\circ-2\angle4\ \ \ ...(\text{iii})$
Now, in $\triangle\text{AEC},\ \ \ \angle\text{AEC}+\angle\text{ECA}+\angle\text{CAE}=180^\circ$ [by angle property sum of a triangle]
$\Rightarrow\angle\text{AEC}=180^\circ-(\angle\text{ECA}+\angle\text{CAE})$
$\Rightarrow\angle\text{AEC}=180^\circ-[(\angle\text{ECO}+\angle\text{OCA})+\angle\text{CAO}+\angle\text{OAE}]$
$=180^\circ-(\angle6+\angle4+\angle4+\angle5)$
$\big[$In $\triangle\text{OCD},\angle6=\angle\text{ECO}$ angles opposite to equal sides are equal$\big]$
$=180^\circ-(2\angle4+\angle5+\angle6)$
$=180^\circ-(180^\circ-\angle\text{AOC}+\angle7+\angle6)[ $From Eq. $(iii)$ and in
$\triangle\text{AOB}.\angle5=\angle7,$ as $($angles opposite to equal sides are equal$)]$
$=\angle\text{AOC}-\frac{\angle3}{2}-\frac{\angle1}{2} [$from Eqs. $(i)$ and $(ii)]$
$=\angle\text{AOC}-\frac{\angle1}{2}-\frac{\angle2}{2}-\frac{\angle3}{2}+\frac{\angle2}{2}$ [adding and subtracting $\frac{\angle2}{2}$]
$=\angle\text{AOC}-\frac{1}{2}(\angle1+\angle2+\angle3)+\frac{\angle8}{2} [ \because\angle2=\angle8$ (vertically opposite angles)]
$=\angle\text{AOC}=\frac{\angle\text{AOC}}{2}+\frac{\angle\text{DOB}}{2}$
$\Rightarrow\angle\text{AEC}=\frac{1}{2}(\angle\text{AOC}+\angle\text{DOB})$
$=\frac{1}{2} [$angle subtended by arc $CXA$ at the centre $+$ angle subtended by arc $DYB$ at the centre$]$
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