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

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To prove : $\angle ABC = \frac{1}{2}(\angle AOC-\angle DOE)$

Construction: Join AC and DE.

Proof :

Let $\angle$ADC = x , $\angle$DOE = y and  $\angle$AOD = z

So,  $\angle$EOC = z (each chord subtends equal angle at centre)

$\angle$AOC + $\angle$DOE +$\angle$AOD + $\angle$EOC = $360 \degree$

$\Rightarrow x+y+z+z=360 \degree$

$\Rightarrow x+y+2z=360 \degree$.........................................1

In $\triangle$ OAD ,

OA = OD  (Radii of the circle)

$\angle$OAD = $\angle$ODA    (angles opposite to equal sides )

$\angle$OAD + $\angle$ODA + $\angle$AOD =$180 \degree$

$\Rightarrow 2\angle OAD+z=180 \degree$

$\Rightarrow 2\angle OAD=180 \degree-z$

$\Rightarrow \angle OAD=\frac{180 \degree-z}{2}$

$\Rightarrow \angle OAD=90 \degree-\frac{z}{2}$.............................................................2

Similarly,

$\Rightarrow \angle OCE=90 \degree-\frac{x}{2}$.............................................................3

$\Rightarrow \angle OED=90 \degree-\frac{y}{2}$..............................................................4

$\angle$ODB is exterior of triangle OAD . So,

$\angle$ODB = $\angle$OAD + $\angle$ODA

$\Rightarrow \angle ODB=90 \degree-\frac{z}{2}+z$                  (from  2)

$\Rightarrow \angle ODB=90 \degree+\frac{z}{2}$.................................................................5

similarly,

$\angle$OBE is exterior of triangle OCE . So,

$\angle$OBE = $\angle$OCE + $\angle$OEC

$\Rightarrow \angle OEB=90 \degree-\frac{z}{2}+z$                  (from  3)

$\Rightarrow \angle OEB=90 \degree+\frac{z}{2}$.................................................................6

From 4,5,6 ;we get

$\angle$BDE = $\angle$BED = $\angle$OEB - $\angle$OED

$\Rightarrow \angle BDE=\angle BED=90 \degree+\frac{z}{2}-(90-\frac{y}{2})=\frac{y+z}{2}$

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

In $\triangle$BDE ,

$\angle$DBE + $\angle$BDE + $\angle$BED = $180 \degree$

$\Rightarrow \angle DBE +y+z=180 \degree$

$\Rightarrow \angle DBE =180 \degree-(y+z)$

$\Rightarrow \angle ABC =180 \degree-(y+z)$...................................................8

Here,  from equation 1,

$\frac{x-y}{2}=\frac{360 \degree-y-2x-y}{2}$

$\Rightarrow \frac{x-y}{2}=\frac{360 \degree-2y-2x}{2}$

$\Rightarrow \frac{x-y}{2}=180 \degree-y-x$...................................9

From 8 and 9,we have

$\angle ABC=\frac{x-y}{2}=\frac{1}{2}(\angle AOC-\angle DOE)$

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