Points A, B, C are on a circle, such that m(arc AB) = m(arc BC) = $120^\circ$. No point, except point B, is common to the arcs. Which is the type of ∆ABC?

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KollegeApply · Accepted Answer

Equilateral triangle Explanation: m(arc AB) = m(arc BC) = 120º Now, m(arc AB) + m(arc BC) + m(arc CA) = 360º ⇒ 120º + 120º + m(arc CA) = 360º ⇒ 240º + m(arc CA) = 360º ⇒ m(arc CA) = 360º – 240º ⇒ m(arc CA) = 120º ∴ m(arc AB) = m(arc BC) = m(arc CA) ⇒ arc AB ≅ arc BC ≅ arc CA ...(Two arcs are congruent if their measures are equal.) ⇒ chord AB ≅ chord BC ≅ chord CA ...(Chords corresponding to congruent arcs of a circle are congruent.) ∴ ∆ABC is an equilateral triangle. ...(All sides of equilateral triangle are equal.) Hence, the correct answer is equilateral triangle.

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