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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.