Question

ABC is an equilateral triangle inscribed in the circle P and Q are two points on the circumference. find angle BPC and angle BQC​

Answer 1

Step-by-step explanation:

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)Read more on Sarthaks.com - https://www.sarthaks.com/978202/abc-is-equilateral-triangle-inscribed-in-circle-is-any-point-on-the-minor-arc-bc-prove-that