\(XY\) and \(XZ\) are two equal chords of a circle. Verify that the bisector of the angle \(YXZ\) passes through the centre of the circle.
Explanation:
\(XY\) and \(XZ\) are two equal chords whose centre is \(O\).
Now, Join \(YZ\).
Draw bisector \(XD\) of \(∠YXZ\)
Therefore, \(∠YXD=∠ZXD\)
In \(△YXO\) and \(△ZXO\),
\(XY = \) (given)
\(∠YXO=∠\) (by construction)
\(XO =\) []
Therefore, \(△YXO≅△ZXO\) []
That is, \(YO = ZO\) [by ] and
\(∠YOX=∠ZOX\) [by ]
Also, \(YO = ZO\) and \(∠YOX=∠ZOX=\)\(^°\)
Thus, \(XO\) is the perpendicular bisector of the chord \(YZ\).
Therefore, the bisector of \(∠YXZ\) passes through the centre \(O\).
That is., \(XD\) passes through the centre \(O\).
Hence proved.