Question.
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Solution:
Let PQ and RS are two equal chords of a given circle and they are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In $\triangle O V T$ and $\triangle O U T$,
OV = OU (Equal chords of a circle are equidistant from the centre)
$\angle O V T=\angle O U T\left(\right.$ Each $\left.90^{\circ}\right)$
OT $=$ OT (Common)
$\therefore \Delta O V T \cong \Delta O U T(R H S$ congruence rule $)$
$\therefore \angle O T V=\angle O T U(B y C P C T)$
Therefore, it is proved that the line joining the point of intersection to the centre makes equal angles with the chords.
Let PQ and RS are two equal chords of a given circle and they are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In $\triangle O V T$ and $\triangle O U T$,
OV = OU (Equal chords of a circle are equidistant from the centre)
$\angle O V T=\angle O U T\left(\right.$ Each $\left.90^{\circ}\right)$
OT $=$ OT (Common)
$\therefore \Delta O V T \cong \Delta O U T(R H S$ congruence rule $)$
$\therefore \angle O T V=\angle O T U(B y C P C T)$
Therefore, it is proved that the line joining the point of intersection to the centre makes equal angles with the chords.