The equidistant point of the triangle is point O , which is shown below.
In the question ,
we have to find the equidistant point in the triangle ,
We know that the equidistant point from the vertices of the triangle is the circumcenter ,
let the circumcenter be point O .
The perpendicular bisectors of the given triangle ABC are AC and BC that intersect at point O .
We have to prove that point O lies on the perpendicular bisector of AB and it is equidistant from A , B and C .
So ,we draw OA ,OB and OC .
we know that any point on perpendicular bisector of a segment is equidistant from endpoints of the segment.
that means , OA = OC and OC = OB .
which means OA = OB .
we also know that any point that is equidistant from end points of a segment lies on its perpendicular bisector.
So, we can say that point O is on the perpendicular bisector of AB .
and
Since OA = OB = OC , the point O is equidistant from A , B and C .
Therefore , the equidistant point of the triangle is the circumcenter (point O) .
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