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.

Given:
RQ and MN are chords of a circle with centre O. MN and RQ intersect at P and MN = RQ.

To prove: \(\angle{OPC}\) = \(\angle{OPB}\)

Construction: Draw \(OC\perp{RQ}\) and \(OB\perp{MN}\). Join OP.

Proof:
In \(\triangle{OCP}\) and \(\triangle{OBP}\),
OC = OB
(Equal chords of a circle are equidistant from the centre)
\(\angle{OCP}\) = \(\angle{OBP}\)
(Each angle is \(90^\circ\))
and OP = OP ...(Common side)

Therefore, by RHS criterion of congruence,
\(\triangle{OCP}\) \(\displaystyle \cong\) \(\triangle{OBP}\)
Also, \(\angle{OPC}\) = \(\angle{OPB}\) ...(By CPCT)