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  • The circumcenter of the triangle ABC is O. Prove that angle OBC plus angle BAC equal to 90 degree.

The circumcenter of the triangle ABC is O. Prove that \angleOBC + \angleBAC = 90º.

Answers (1)

Given that the circumcenter of the triangle ABC is O. So O will be the radius of the circle passing through the points A, B, C

Now, OB = OC = radius of the circle

We know that angles opposite to equal sides in a triangle are equal

\Rightarrow \angleOBC = \angleOCB = Let x

In DOBC,

\therefore x + x + \angleBOC = 180° (angle sum property of a triangle)

2x + \angleBOC = 180°

\angleBOC = 180° – 2x

Now, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle

So considering the arc BC,

\angleBOC = 2 \angleBAC

Putting the value of \angleBOC

180° – 2x = 2 \angleBAC

90° – x = \angleBAC

We have to find  \angleOBC + \angleBAC

\therefore \angleÐBAC +\angleOBC = (90° – x) + x

\angleBAC + \angleOBC = 90°

Hence proved

Posted by

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