Question

Prove :
The angle bisectors of a triangle are concurrent at a point equidistant from the sides of the triangle.

Answer 1

Three perpendicular bisectors of a triangle sides are concurrent, in other words, they intersect at one point.
This intersection point is equidistant from the three triangle vertices and is the center of the circumscribed circle of the triangle.

Proof
Figure 1 shows the triangle ABC with the midpoints D, E and F
of its three sides AB, BC and AC respectively. The straight lines
DG, EH and FI are perpendicular to the sides AB, BC and AC,
and are passing through midpoints D, E and F.

Since the straight lines DG and EH are perpendicular to the
sides AB and BC respectively, they can not be parallel,
otherwise the sides AB and BC would be in one straight line
what is not the case. Therefore, the straight lines DG and
EH intersect in some point P.

From the lesson A perpendicular bisector of a segment
(Theorem 1) we know that the points of the perpendicular DG
are equidistant from the endpoints A and B of the segment AB.

Figure 1. To the Theorem

Figure 2. To the proof of the Theorem
In particular, the point P is equidistant from the endpoints A and B. This means that the straight segments AP and BP (Figure 2) are of equal length: AP = BP.

By the same reason, the points of the perpendicular EH are equidistant from the endpoints B and C of the segment BC.
In particular, the point P is equidistant from the endpoints B and C. This means that the straight segments BP and CP (Figure 2) are of equal length: BP = CP.

Two equalities above imply that the straight segments AP and CP are of equal length too: AP = CP. In other words, the point P is equidistant from the points A and C.
In turn, it implies that the intersection point P lies at the midpoint perpendicular GI to the segment AC in accordance to the Theorem 2 of the lesson
A perpendicular bisector of a segment. In other words, the midpoint perpendicular GI passes through the point P.

Thus, we have proved that all three midpoint perpendiculars DG, EH and FI pass through the point P and have this point as their common intersection point.
Since the point P is equidistant from the triangle vertices A, B and C, it is the center of the circumscribed circle of the triangle ABC (Figure 2).
So, all the statements of the Theorem are proved.

The proved property provides the way of constructing a circumscribed circle for a given triangle.
To construct the center of such a circle, it is enough to construct the midpoints of any two the triangle sides and then to construct the midpoint perpendiculars to these triangle sides.
The intersection point of the midpoint perpendiculars is the center of the circumscribed circle of the triangle.
The radius of the circumscribed circle is equal to the distance from the constructed center of the circumscribed circle to any of the triangle vertex.
All these operations can be done with the use of a ruler and a compass as it is explained in the lesson How to bisect a segment using a compass and a ruler under the current
topic Triangles of the section Geometry in this site.

Summary
Three perpendicular bisectors of a triangle sides are concurrent, in other words, they intersect at one point.
This intersection point is equidistant from the three triangle vertices and is the center of the circumscribed circle of the triangle.

For a given triangle, the circumscribed circle can be constructed with the use of a ruler and a compass as follows:
- First construct the midpoints for any two the triangle sides and then construct the midpoint perpendiculars to these triangle sides.
The intersection point of the midpoint perpendiculars is the center of the circumscribed circle of the triangle.
- The radius of the circumscribed circle is equal to the distance from the constructed center of the circumscribed circle to any the triangle vertex.

Angle bisectors of a triangle, altitudes of a triangle and medians of a triangle have the similar properies:
- angle bisectors of a triangle are concurrent;
- altitudes of a triangle are concurrent;
- medians of a triangle are concurrent.

Answer 2

it may help u dear

Answer 3

is this geometry?