Question

$ABC$ is a triangle. Locate a point in the interior of $△ABC$ which is equidistant from all the vertices of $△ABC$.

Answer 1

Let $OD$ and $OE$ be the perpendicular bisectors of sides $BC$ and $AB$ of $△ABC$ respectively.

$\therefore$ By perpendicular bisector theorem,

$O$ is equidistant from the end points of seg $BC$ i.e. points $B$ and $C$.

Similarly, point $O$ is equidistant from end points of seg $AC$ i.e points $C$ and $A.$

Hence, the point of intersection $O$ of the perpendicular bisectors of sides $AB$ and $BC$ is equidistant from vertices $A,B,C$ of $△ABC.$