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

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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)

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