 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.