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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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M mansi

Given : AD = CE

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

Construction: Join AC and DE.

Proof : 

            

Let \angleADC = x , \angleDOE = y and  \angleAOD = z

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

   \angleAOC + \angleDOE +\angleAOD + \angleEOC = 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)

       \angleOAD = \angleODA    (angles opposite to equal sides )

    \angleOAD + \angleODA + \angleAOD =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

\angleODB is exterior of triangle OAD . So,

 \angleODB = \angleOAD + \angleODA

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

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

similarly,

\angleOBE is exterior of triangle OCE . So,

 \angleOBE = \angleOCE + \angleOEC

\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

            \angleBDE = \angleBED = \angleOEB - \angleOED

\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 \triangleBDE , 

         \angleDBE + \angleBDE + \angleBED = 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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