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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)