2. If two equal chords of a circle intersect within the circle, prove that the segments of
one chord are equal to corresponding segments of the other chord.

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

To prove: MP = PB and PN = AP

Construction: Draw \(OD\perp{MN}\) and \(OC\perp{AB}\).Join OP

Proof: DM = DN = (1\2) MN ...(Since, Perpendicular from centre bisects the chord)

Similarly, AC = CB = (1\2) AB ...(Since, Perpendicular from centre bisects the chord)

Therefore, we get,

MD = BC and DN = AC ... (i)(Since, MN = AB)

In \(\triangle{ODP}\) and \(\triangle{OPC}\),

OD = OC ...(Equal chords of a circle are equidistant from the centre)

\(\angle{ODP}\) = \(\angle{OCP}\)

and OP = OP ...(Common side)

Therefore, by RHS criterion of congruence,

\(\triangle{ODP}\) \(\displaystyle \cong\) \(\triangle{OPC}\)

Also, DP = PC ...(By CPCT) ...(ii)

On adding Equations (i) and (ii), we get,

MD + DP = BC + PC

i.e., MP = PB

Also, On subtracting Equation (ii) and (i),we get,

DN + DP = AC + PC

i.e., PN = AP

Hence, MP = PB and PN = AP are proved.