# An equilateral triangle ABC is inscribed in a circle. P is any point on the minor arc BC. Prove that PA = PB + PC.

Equilateral ΔABC inscribed in a circle C(O, r) with point P on minor arc BC.

To Prove: PA = PB + PC
Construction: Produce BP to Q such that PQ = PC. Join C to Q.
Proof: ABPC is a cyclic quadrilateral  ... ∠A + ∠BPC = 180°               .............. (i)
But ∠A = 60° [angle of a Equilateral triangle]
... ∠BPC = 180° - 60° = 120°
... ∠CPQ = 180° - ∠BPC
= 180° - 120°
= 60°                .............. (ii)
Now in ΔPQC,
∠3 + ∠4 + ∠Q = 180°
... ∠3 + ∠Q = 120°
But ∠3 = ∠Q
... ∠3 = ∠Q = 60°
Hence, ΔPQC is equilateral
∴ ∠2 = ∠3 = 60°
∴ ∠2 + ∠BCP = ∠3 + ∠BCP
⇒ ∠ACP = ∠BCQ
Now in ΔACP and ΔBCQ, ∠1 = ∠5 [Angles in the same segment of the circle]
∠ACP = ∠BCQ [proved above]
and PC = QC [by construction]
... ΔACP @ DBCQ
... AP = BQ = BP + PQ
⇒ AP = BP + PC [PC = PQ].

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