Question

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

Answer 1

Step-by-step explanation:

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

∴ 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.