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Question:

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

Solution:

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.