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# 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 = ($$\frac{1}{2}$$ ) MN
(Since, Perpendicular from centre bisects the chord)
Similarly,
AC = CB = ($$\frac{1}{2}$$ ) 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
$$\Rightarrow$$ MP = PB
Also, On subtracting Equation (ii) and (i),we get,

DN + DP = AC + PC
$$\Rightarrow$$ PN = AP

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