Given: Point C is the centre of the circle.
chord PM ≅ chord PN
To prove: Ray PC is the bisector of ∠NPM.
Construction: Draw seg CR ⊥ chord PN,
P-R-N
seg CQ ⊥ chord PM, P-Q-M
Proof:
chord PM chord PN [Given]
seg CR ⊥ chord PN
seg CQ ⊥ chord PM [Construction]
∴ seg CR ≅ seg CQ ….(i) [Congruent chords are equidistant from the centre]
In ∆PRC and ∆PQC, ∠PRC ≅ ∠PQC [Each is of 90°]
seg CR ≅ seg CQ [From (i)]
seg PC ≅ seg PC [Common side]
∴ ∆PRC ≅ ∆PQC [Hypotenuse side test]
∴ ∠RPC ≅ ∠QPC [c. a. c. t.]
∴ ∠NPC ≅ ∠MPC [N- R-P, M-Q-P]
∴ Ray PC is the bisector of ∠NPM.
