Seg PM and seg PN are congruent chords of a circle with center C. Show that the ray PC is the bisector of ∠NPM.

Given that PM = PN

We know that Congruent chords of a circle are equidistant from the center of the circle.

Therefore, AC = CB ………………(1)

Also,

A perpendicular drawn from the centre of a circle on its chord bisects

the chord.

CB bisects PN as PB = BN,

Similarly, CA bisects PM as PA = AM.

In ΔAPC and ΔBPC,

∠CAP = ∠ CBP = 90°

PC = PC (common side)

AC = CB (From eq (1))

∴ ΔAPC≅ ΔBPC (RHS congruence)

∴ ∠ APC = ∠ BPC (by CPCT)

Hence proved that PC is the bisector of ∠ NPM.