Answer :

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)

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