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The inscribed circle of a triangle
If we construct the interior angle bisectors of a triangle, then it seems like the three rays would intersect each other at one point. We cannot say it for sure based on sketching since it is possible that our construction is inaccurate, but it is definitely worth examining this question a bit more deeply.
Let us consider triangle ABC, and let us draw the interior angle bisectors starting from vertices A and B (Figure 69). These definitely intersect each other at an internal point M of the triangle, or else there would be a \latex{180^{\circ}} interior angle at both A and B.
M lies on the interior angle bisector starting from vertex A, therefore it is equidistant from the straight line containing side AB and the straight line containing side AC, i.e.
\latex{d(M; AB)=d(M; AC)}. (1)
M also lies on the interior angle bisector starting from vertex B, therefore\latex{d(M; AB)=d(M; BC)}. (2)
Figure 69
\latex{ M }
\latex{ A }
\latex{ B }
\latex{ C }
The combination of equalities (1) and (2) implies that
\latex{d(M; AC)=d(M; BC)},
i.e. M is equidistant from the straight lines containing side AC and side BC respectively. Since the third angle bisector is the set of those points which are equidistant from side AC and side BC, therefore M also lies on the interior angle bisector starting from vertex C. Thus \latex{M} is a point of all three interior angle bisectors, so we have proven the assumption formulated apropos of the construction.
proof
Figure 70
\latex{ M }
\latex{ A }
\latex{ B }
\latex{ C }
\latex{ r }
\latex{ r }
\latex{ r }
THEOREM: The three interior angle bisectors of a triangle intersect each other at one point.
Point M is equidistant from all three straight lines containing the sides, therefore a circle can be constructed about point M as the centre, which touches all three sides of the triangle at an internal point. This circle is the inscribed circle (or incircle) of the triangle. (Figure 70)
THEOREM: The intersection point of the interior angle bisectors of a triangle is the centre of the inscribed circle.
Example 1
Let us find the points of the plane of the triangle which are equidistant from all three straight lines containing the sides.
Solution
We have seen that there is exactly one such point inside the triangle: the intersection point of the interior angle bisectors of the triangle, centre O of the inscribed circle of the triangle. If we take the straight lines of the angle bisectors of the interior and exterior angles of the triangle, then it can be shown similarly to the method seen for the interior angle bisectors that the straight line of the angle bisector of one interior angle and the straight lines of angle bisectors of the exterior angles belonging to the other two vertices of the triangle also intersect each other at one point. It means three new suitable points outside the triangle, let them be denoted by \latex{O_1}, \latex{O_2}, \latex{O_3} according to figure 71. All these points are equidistant from the straight lines of all three sides, so circles can be drawn about them which touch one side of the triangle at an internal point and also touch the straight lines of the other two sides at external points respectively.
Figure 71
\latex{O_1}
\latex{O_2}
\latex{O_3}
These three circles are called the escribed circles (or excircles) of the triangle.
Example 2
Let \latex{P} denote the perimeter of a triangle, and let \latex{r} denote the radius of the inscribed circle. Let us prove that the area of the triangle can be given with formula \latex{A=\frac{P}{2}\times r}.
Figure 72
\latex{ B }
\latex{ C }
\latex{ A }
\latex{ O }
\latex{ c }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ b }
\latex{ a }
Solution
Let O denote the centre of the inscribed circle, and let \latex{a, b, c } denote the length of the sides of the triangle (Figure 72). Line segments AO, BO, CO divide the triangle into three smaller triangles. The sum of the areas of these small triangles gives the area of the original triangle, i.e.
\latex{A_{ABC}=\frac{a\times r}{2}+\frac{b\times r}{2}+\frac{c\times r}{2}=\left(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)\times r= \frac{a+b+c}{2}\times r= \frac{P}{2}\times r.}
With this we have proven the statement. Half of the perimeter (semiperimeter) of a triangle is usually denoted by s. Using this the relation we have just proven is as follows:
\latex{A=s\times r}.
\latex{A=s\times r},
where s is half the perimeter
of the triangle and r is the
radius of the inscribed circle.
of the triangle and r is the
radius of the inscribed circle.
Exercises
{{exercise_number}}. Find the angles of the isosceles triangle if the angle included by the interior angle bisectors starting from the end-points of its base is
- \latex{120^{\circ};}
- \latex{106^{\circ};}
- \latex{96^{\circ};}
- \latex{160^{\circ}}.
{{exercise_number}}. Construct the inscribed circle of triangle ABC if
- \latex{AB= 6 \;cm, \; BC=5\;cm,\; CA=4\;cm;}
- \latex{AB= 4\;cm, \; \alpha=60^{\circ},\; \beta=45^{\circ};}
- \latex{AB= 4\;cm, \; BC=6.5\; cm,\; \beta= 30^{\circ};}
- \latex{AB= 5.5\;cm, \; BC=4\; cm,\; \gamma= 75^{\circ}}.
{{exercise_number}}. Construct the escribed circles of the triangles determined in the previous exercise.
{{exercise_number}}. Calculate the area of the triangle if denoting its perimeter by P and the radius of its inscribed circle by r:
- \latex{P= 20 \;cm, \; r=5\;cm}
- \latex{P= 17 \;cm, \; r=2.5\;cm}
- \latex{P= 2\;m, \; r=6.5\; cm}
- \latex{P= 48.3\;cm, \; r=6.8\; cm}