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  • Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Q 4. Let the vertex of an angle $A B C$ be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that $\angle A B C$ is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answers (1)

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Given : $\mathrm{AD}=\mathrm{CE}$
To prove :
$\angle A B C=\frac{1}{2}(\angle A O C-\angle D O E)$
Construction: Join AC and DE.
Proof:

Let $\angle \mathrm{ADC}=\mathrm{x}, \angle \mathrm{DOE}=\mathrm{y}$ and $\angle \mathrm{AOD}=\mathrm{z}$
So, $\angle E O C=z$ (each chord subtends an equal angle at the centre)
$\angle \mathrm{AOC}+\angle \mathrm{DOE}+\angle \mathrm{AOD}+\angle \mathrm{EOC}=360^{\circ}$
$\Rightarrow x+y+z+z=360^{\circ}$
$\Rightarrow x+y+2 z=360^{\circ}$---------(1)
$\text { In } \triangle \mathrm{OAD}$,
$\mathrm{OA}=\mathrm{OD}$ (Radii of the circle)
$\angle \mathrm{OAD}=\angle \mathrm{ODA}$(angles opposite to equal sides)

$\angle \mathrm{OAD}+\angle \mathrm{ODA}+\angle \mathrm{AOD}=180^{\circ}$
$\Rightarrow 2 \angle O A D+z=180^{\circ}$
$\Rightarrow 2 \angle O A D=180^{\circ}-z$
$\Rightarrow \angle O A D=\frac{180^{\circ}-z}{2}$
$\Rightarrow \angle O A D=90^{\circ}-\frac{z}{2}$-------(2)

Similarly,

$\Rightarrow \angle O C E=90^{\circ}-\frac{x}{2}$------(3)
$\Rightarrow \angle O E D=90^{\circ}-\frac{y}{2}$------(4)
$\angle O D B$ is the exterior of triangle OAD. 
So, $\angle \mathrm{ODB}=\angle \mathrm{OAD}+\angle \mathrm{ODA}$
$\Rightarrow \angle O D B=90^{\circ}-\frac{z}{2}+z$   (from 2)
$\Rightarrow \angle O D B=90^{\circ}+\frac{z}{2}$-------(5)

Similarly,

$\angle O B E$ is the exterior of triangle OCE. 
So, $\angle \mathrm{OBE}=\angle \mathrm{OCE}+\angle \mathrm{OEC}$
$\Rightarrow \angle O E B=90^{\circ}-\frac{z}{2}+z$    (from 3)
$\Rightarrow \angle O E B=90^{\circ}+\frac{z}{2}$------(6)
From 4,5,6, we get
$\angle \mathrm{BDE}=\angle \mathrm{BED}=\angle \mathrm{OEB}-\angle \mathrm{OED}$
$\Rightarrow \angle B D E=\angle B E D=90^{\circ}+\frac{z}{2}-\left(90-\frac{y}{2}\right)=\frac{y+z}{2}$

\Rightarrow \angle BDE+\angle BED=y+z..................................................7

$\Rightarrow \angle B D E+\angle B E D=y+z$---------(7)
In $\triangle \mathrm{BDE}$
$\angle \mathrm{DBE}+\angle \mathrm{BDE}+\angle \mathrm{BED}=180^{\circ}$
$\Rightarrow \angle D B E+y+z=180^{\circ}$
$\Rightarrow \angle D B E=180^{\circ}-(y+z)$
$\Rightarrow \angle A B C=180^{\circ}-(y+z)$--------(8)
Here, from equation 1 ,
$\frac{x-y}{2}=\frac{360^{\circ}-y-2 x-y}{2}$

$\Rightarrow \frac{x-y}{2}=\frac{360^{\circ}-2 y-2 x}{2}$
$\Rightarrow \frac{x-y}{2}=180^{\circ}-y-x$-------(9)
From 8 and 9, we have
$\angle A B C=\frac{x-y}{2}=\frac{1}{2}(\angle A O C-\angle D O E)$

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