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.