Given, ABC is an equilateral triangle and P is a point on the minor arc BC.
∠ABC = ∠BAC = ∠BCA = 60°
Let ∠BCP = x
Produce BP to Q such that PQ = PC. Join CQ.
∠CPQ isthe external angle ofthe cyclic quadrilateral ABPC.
∵ ∠CPQ = ∠BAC = 60°.
∵ PC = PQ, and ∠CPQ = 60°, therefore ∆ CPQ is equilateral.
Consider the triangles ACP and BCQ.
∠ACP = 60 + x, ∠BCQ = 60 + x
Now in Δs ACP and BCQ
∠ACP = ∠BCQ = 60 + x (Proved)
∠CAP = ∠CBP (∠CBQ) (Angles in the same segment PC)
AC = BC (sides of equilateral ∆ ABC)
∴ ∆ ACP ≅ ∆ BCQ (ASA)
⇒ AP = BQ
⇒ AP = BP + PQ
⇒ AP = BP + PC (∵ PC = PQ)
