Demonstrate that the bisector of the angle formed by two equal chords starting from the same point on a circle must pass through the center of the circle.
Explanation:
\(AB\) and \(AC\) are two equal chords whose centre is \(O\).
Now, Join \(BC\).
Draw bisector \(AD\) of \(∠BAC\)
Therefore, \(∠BAD=∠CAD\)
In \(△BAO\) and \(△CAO\),
\(AB = \) (given)
\(∠BAO=∠\) (by construction)
\(AO =\) []
Therefore, \(△BAO≅△CAO\) []
That is, \(BO = CO\) [by ] and
\(∠BOA=∠COA\) [by ]
Also, \(BO = CO\) and \(∠BOA=∠COA=\)\(^°\)
Thus, \(AO\) is the perpendicular bisector of the chord \(BC\).
Therefore, the bisector of \(∠BAC\) passes through the centre \(O\).
That is., \(AD\) passes through the centre \(O\).
Hence proved.