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

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 linewhat 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 TheoremIn 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. 

The answer would be option 4