$ \mathrm{ABC} $ is a triangle. Locate a point in the interior of $ \triangle \mathrm{ABC} $ which is equidistant from all the vertices of $ \triangle \mathrm{ABC} $.

Given:
 $ABC$ is a triangle.

To do:

We have to locate a point in the interior of $\triangle ABC$ which is equidistant from all the vertices of $\triangle ABC$.

Solution:

Let us consider a $\triangle ABC$

We know that,

A point in the interior of the triangle which is equidistant from all the vertices is called the circumcentre of the triangle.

We also know that,

The circumcenter of the triangle is the point where all the perpendicular bisectors of the sides of the triangle intersect.

Therefore,

In $\triangle ABC$ let us draw three perpendicular bisectors from points $A$, $B$ and $C$ and 

Let us mark the point of intersection as point $O$

Therefore, $O$ is the required point in the interior of $\triangle ABC$ which is equidistant from all the vertices of $\triangle ABC$.

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Updated on: 10-Oct-2022

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