Let the vertex of an angle \( \mathrm{ABC} \) be located outside a circle and let the sides of the angle intersect equal chords \( \mathrm{AD} \) and \( \mathrm{CE} \) with the circle. Prove that \( \angle \mathrm{ABC} \) is equal to half the difference of the angles subtended by the chords \( A C \) and \( D E \) at the centre.
Given:
Let the vertex of an angle \( \mathrm{ABC} \) be located outside a circle and let the sides of the angle intersect equal chords \( \mathrm{AD} \) and \( \mathrm{CE} \) with the circle.
To do:
We have to prove that \( \angle \mathrm{ABC} \) is equal to half the difference of the angles subtended by the chords \( A C \) and \( D E \) at the centre.
Solution:
$AD = CE$
We know that,
An exterior angle of a triangle is equal to the sum of interior opposite angles.
This implies, in $\triangle BAE$,
$\angle DAE = \angle ABC+\angle AEC$........(i)
$DE$ subtends $\angle DOE$ at the centre and $\angle DAE$ in the remaining part of the circle.
This implies,
$\angle DAE = \frac{1}{2}\angle DOE$.........(ii)
Similarly,
$\angle AEC = \frac{1}{2}\angle AOC$..........(iii)
From (i), (ii) and (iii), we get,
$\frac{1}{2}\angle DOE = \angle ABC+\frac{1}{2}\angle AOC$
$\angle ABC =\frac{1}{2}(\angle DOE-\angle AOC)$
Hence proved.
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