Question

1. ABC is a triangle. Locate a point in the interior of A ABC which is equidistant from all
the vertices of A ABC.
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Answer 1

Answer:

let the point be P, and virtices be A, B, C

given AP=AB=AC(circradius)

circumcentre is the point that is equidistant from all vertices it is also Intersection of the peependicular bosectors.

hope it helps u