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

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

Thus, draw perpendicular bisectors of 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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