Question

prove that point equidistant from vertices of triangle is called circumcentre​

Answer 1

Answer:

The three perpendicular bisectors of a triangle meet in a single point, called the circumcenter .

Circumcenter Theorem

The vertices of a triangle are equidistant from the circumcenter.

Given:

ΔABC , the perpendicular bisectors of AB¯¯¯¯¯,BC¯¯¯¯¯ and AC¯¯¯¯¯ .

To prove:

The perpendicular bisectors intersect in a point and that point is equidistant from the vertices.

The perpendicular bisectors of AC¯¯¯¯¯ and BC¯¯¯¯¯ intersect at point O .

Let us prove that point O lies on the perpendicular bisector of AB¯¯¯¯¯ and it is equidistant from A , B and C .

Draw OA¯¯¯¯¯,OB¯¯¯¯¯ and OC¯¯¯¯¯ .

Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.

So, OA=OC and OC=OB .

By the transitive property,

OA=OB .

Any point equidistant from the end points of a segment lies on its perpendicular bisector.

So, O is on the perpendicular bisector of AB¯¯¯¯¯ .

Since OA=OB=OC , point O is equidistant from A , B and C .

This means that there is a circle having its center at the circumcenter and passing through all three vertices of the triangle.  This circle is called the circumcircle .

Explanation:

Answer 2

Explanation:

The CIRCUMCENTER of a triangle is the point in the plane equidistant from the three vertices of the triangle. ... The circumcenter was constructed by identifying the midpoints of the segments AC, CD, and DA. Then a perpendicular line was drawn through the midpoints perpendicular to the side segment