# Lesson Angle bisectors of a triangle are concurrent

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## Angle bisectors of a triangle are concurrent

In this lesson we consider the angle bisectors of a triangle. These bisectors possess a remarkable property: all three intersect at one point.

The property is proved in this lesson. The proof is based on the angle bisector properties that were proved in the lesson An angle bisector properties under the current
topic Triangles of the section Geometry in this site.

Theorem
Three angle bisectors of a triangle are concurrent, in other words, they intersect at one point.
This intersection point is equidistant from the three triangle sides and is the center of the inscribed circle of the triangle.
 Proof Figure 1 shows the triangle ABC with the angle bisectors AD,           BE and CF of its three angles A, B and C respectively. The points D, E and F are the intersection points of the angle bisectors and the opposite triangle sides. Since the straight lines AD and BE are the angle bisectors to the angles A and B 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 AD and BE intersect in some point P. From the lesson An angle bisector properties (Theorem 1) we know that the points of the angle bisector AD are equidistant from the sides AB and AC of the angle BAC. Figure 1. To the Theorem Figure 2. To the proof of the Theorem
In particular, the point P is equidistant from the sides AB and AC of the angle BAC. This means that the perpendiculars GP and HP (Figure 2) drawn from the point P to
the sides AB and AC are of equal length: GP = HP.

By the same reason, the points of the angle bisector BE are equidistant from the sides AB and BC of the angle ABC. In particular, the point P is equidistant from
the sides AB and BC of the angle ABC. This means that the perpendiculars GP and IP (Figure 2) drawn from the point P to the sides AB and BC are of equal length: GP = IP.

Two equalities above imply that the perpendiculars HP and IP are of equal length too: HP = IP. In other words, the point P is equidistant from the sides AC and BC of the
angle ACB. In turn, it implies that the intersection point P lies at the angle bisector CF of the angle ACB in accordance to the Theorem 2 of the lesson
An angle bisector properties. In other words, the angle bisector CF of the angle ACB passes through the point P.

Thus, we have proved that all three angle bisectors 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 sides AB, BC and AC, it is the center of the inscribed circle of the triangle ABC (Figure 2).
So, all the statements of the Theorem are proved.

The proved property provides the way of constructing an inscribed circle for a given triangle.
To find the center of such a circle, it is enough to construct the angle bisectors of any two triangle angles and identify their intersection point. This intersection point
is the center of the inscribed circle of the triangle. To get the radius of the inscribed circle you should to construct the perpendicular from the found center of the
inscribed circle to any triangle side.
All these operations can be done with the use of a ruler and a compass as it is explained in the lessons
How to bisect an angle using a compass and a ruler and
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 angle bisectors of a triangle are concurrent, in other words, they intersect at one point.
This intersection point is equidistant from the three triangle sides and is the center of the inscribed circle of the triangle
.

For a given triangle, the inscribed circle can be constructed with the use of a ruler and a compass as follows:
- First construct the angle bisectors for any two the triangle angles.
The intersection point of the angle bisectors is the center of the inscribed circle of the triangle.
- The radius of the inscribed circle is equal to the distance from the constructed center of the inscribed circle to any the triangle side.
To get the radius of the inscribed circle of the triangle, construct the perpendicular from the found center to any the triangle side
.

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

These properties are proved in the lessons
Perpendicular bisectors of a triangle are concurrent;
Altitudes of a triangle are concurrent;
Medians of a triangle are concurrent
that are under the current topic Triangles of the section Geometry in this site.

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