A kosarad üres
The circumscribed circle of a triangle
If we construct the perpendicular bisectors of the sides of an acute triangle, of a right-angled triangle and of an obtuse triangle, then based on the figures we might get the assumption that in all three cases these straight lines intersect each other at one point.
If our construction is precise enough, then it is also obvious that this common point is inside the triangle in the case of an acute triangle, on the hypotenuse in the case of a right-angled triangle, and outside the triangle in the case of an obtuse triangle.
Similarly to the case of the interior angle bisectors we are going to prove again that our observations are reasonable.
Let us take the perpendicular bisectors of side AB and side BC of triangle ABC, let them be denoted by \latex{f_{AB}} and \latex{f_{BC}}. (Figure 73)
\latex{f_{AB}} and \latex{f_{BC}} intersect each other at one point for sure, otherwise there would be a \latex{180^{\circ}} interior angle at vertex B of the triangle.
Let us take the perpendicular bisectors of side AB and side BC of triangle ABC, let them be denoted by \latex{f_{AB}} and \latex{f_{BC}}. (Figure 73)
\latex{f_{AB}} and \latex{f_{BC}} intersect each other at one point for sure, otherwise there would be a \latex{180^{\circ}} interior angle at vertex B of the triangle.
Let \latex{f_{AB}\cap f_{BC}=M}. \latex{M\in f_{AB}}, which implies that
\latex{AM=BM}. (1)
\latex{M\in f_{BC}}, therefore\latex{BM=CM}. (2)
The combination of (1) and (2) implies that\latex{AM=CM}.
Since the perpendicular bisector of AC is the set of those points which are equidistant from vertex A and vertex C, therefore M also lies on the perpendicular bisector of side AC, i.e. \latex{M\in f_{AC}}.
With this we have proven our assumption.
With this we have proven our assumption.
proof
Figure 73
\latex{f_{AB}}
\latex{f_{BC}}
\latex{ C }
\latex{ B }
\latex{ A }
\latex{ M }
THEOREM: The three perpendicular bisectors of the sides of a triangle intersect each other at one point.
In our proof we did not assume anything about the shape of the triangle, thus the above theorem is valid for every triangle.
Point M is equidistant from all three vertices of the triangle, therefore a circle can be constructed about point M as the centre, which passes through all three vertices of the triangle.
This circle is the circumscribed circle (or circumcircle) of the triangle. (Figure 74)
\latex{MA=MB=MC=R}
Figure 74
\latex{ M }
\latex{ R }
\latex{ R }
\latex{ R }
\latex{ B }
\latex{ A }
\latex{ C }
THEOREM: The intersection point of the perpendicular bisectors of the sides of a triangle is the centre of the circumscribed circle.
The centre of the circumscribed circle is inside the triangle in the case of an acute triangle, and is outside the triangle in the case of an obtuse triangle. (Figure 75)
Figure 75
We are going to come back to the case of right-angled triangles in the next chapter.
Example 1
Let us prove that any acute triangle can be divided into three isosceles triangles.
Solution
Let us consider the centre of the circumscribed circle of the triangle. Connecting it with the vertices we get a suitable division, triangles ABO, BCO and CAO are isosceles triangles. (Figure 76)
Figure 76
\latex{ R }
\latex{ R }
\latex{ O }
\latex{ R }
\latex{ B }
\latex{ A }
\latex{ C }
Example 2
Three non-collinear points are given in the plane. Let us give a fourth point so that the four points lie on one circle.
Solution
The three given points determine a triangle. We can construct the circumscribed circle of this triangle. Any point of the circumscribed circle (other than the given ones) can be this fourth point.
Exercises
{{exercise_number}}. Construct the circumscribed 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=4\;cm, BC= 4\;cm, \gamma=75^{\circ}}.
{{exercise_number}}. Construct an isosceles triangle if the radius of its circumscribed circle and the length of
- the base are given;
- the leg are given.
Examine the condition of solvability in both cases.
