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Answer :

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.

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