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1. ABC is a triangle. Locate a point in the interior of \small \Delta ABC which is equidistant from all the vertices of \small \Delta ABC.

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We know that the circumcenter of a triangle is equidistant from all the vertices. Also, the circumcenter is the point of intersection of the perpendicular bisectors of the sides of a triangle.

Thus, draw perpendicular bisectors on each side of the triangle ABC. And let them meet at a point, say O. 

Hence, O is the required point, which is equidistant from all the vertices.

 

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Devendra Khairwa

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