It's more easy than you think

Geometry Level 2

In the figure, $A$ and $B$ are on the circumference of the circle with center $O$ . $OA$ is parallel to $BC$ and $\angle ABO=\dfrac{\pi}{3}$ . What is $\angle BOC$ ?

\frac{2\pi}{3}

\frac{\pi}{3}

\frac{7\pi}{18}

\frac{5\pi}{9}

\frac{4\pi}{9}

2 solutions

Chew-Seong Cheong
Oct 9, 2019

Since $OA$ and $OB$ are radius of the circle, $OA=OB$ and $\triangle ABO$ is an isosceles triangle. This means that $\angle BAO = \angle ABO = \dfrac \pi 3$ . Then $\angle AOB = \pi - \dfrac \pi 3 - \dfrac \pi 3 = \dfrac \pi 3$ , implying $\triangle ABO$ is equilateral. Since $OA\ ||\ BC$ , $\implies \angle AOB = OBC = \dfrac \pi 3$ . Again $OB=OC$ , as they are radius of the circle and $\triangle BCO$ is equilateral and $\angle BOC = \boxed{\frac \pi 3}$ .

@Fahim Muhtamim , you have to mention that the figure is a circle and $O$ is its center. We need a space after comma (,). Use proper English "and" instead of "&". I will amend the problem wording for you.

Chew-Seong Cheong - 1 year, 8 months ago

A Former Brilliant Member
Oct 9, 2019

$\triangle AOB$ is equilateral. So $\angle AOB=\dfrac{π}{3}$ . Being alternate angles, $\angle AOB$ and $\angle OBC$ are equal. So $\triangle OBC$ is also equilateral. Therefore $\angle BOC$ is $\dfrac{π}{3}$

It's more easy than you think

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