Points A, B, C are on a circle, such that m(arc AB) = m(arc BC) =…

MCQ

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

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Equilateral triangle
B
Scalene triangle
C
Right angled triangle
D
Isosceles triangle

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Answer

Correct option: A.

Equilateral triangle

$m(\operatorname{arc} A B)=m(\operatorname{arc~BC})=120^{\circ}$
Now,
$m(\operatorname{arc} A B)+m(\operatorname{arc} B C)+m(\operatorname{arc} C A)=360^{\circ}$
$\Rightarrow 120^{\circ}+120^{\circ}+m(\operatorname{arc} C A)=360^{\circ}$
$\Rightarrow 240^{\circ}+m(\operatorname{arc} C A)=360^{\circ}$
$\Rightarrow m(\operatorname{arc} C A)=360^{\circ}-240^{\circ}=120^{\circ}$
$\therefore m(\operatorname{arc} A B)=m(\operatorname{arc} B C)=m(\operatorname{arc} C A)$
$\Rightarrow \operatorname{arc} A B \cong \operatorname{arc} B C \cong \operatorname{arc} C A$
(Two arcs are congruent if their measures are equal)
$\Rightarrow$ chord $A B \cong$ chord $B C \cong$ chord $C A$ (Chords corresponding to congruent arcs of a circle are congruent)
$\therefore \triangle \mathrm{ABC}$ is an equilateral triangle. ..(All sides of equilateral triangle are equal)
Hence, the correct answer is option Equilateral triangle.