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Thales’ theorem and some of its applications
Let us take a circle with centre O and one of its diameters: AB. Let us consider an arbitrary point C of the circle not coinciding with point A and B (Figure 77). The taken circle is the circumscribed circle of triangle ABC, and its centre is the mid-point of side AB.
Figure 77
\latex{\beta}
\latex{\alpha}
\latex{\alpha}
\latex{\beta}
\latex{ O }
\latex{ B }
\latex{ C }
\latex{ A }
Let us examine triangle ABC. Let us connect O with C. As AO = OC, therefore
triangle AOC is an isosceles triangle. It implies that
triangle AOC is an isosceles triangle. It implies that
\latex{\angle CAO= \angle OCA = \alpha}.
Similarly we get that triangle COB is also an isosceles triangle, which implies
that
that
\latex{\angle BCO = \angle OBC = \beta}.
Let us write the sum of the interior angles of triangle ABC.
\latex{\angle CAB + \angle ABC + \angle BCA = \alpha + \beta + (\alpha + \beta) = 180^{\circ}},
i.e.
\latex{2(\alpha + \beta) = 180^{\circ}}, which implies that \latex{\alpha + \beta = 90^{\circ}}.
We got that the interior angle at vertex C of triangle ABC is a right angle.
proof
The theorem we have just proven dates back to the 6th century BC, the first known Greek mathematician, THALES discovered it.
THALES’ THEOREM: If we connect the two end-points of any of the diameters of a circle with any other point of the circle, then we get a right-angled triangle the hypotenuse of which is the diameter of the circle.
Thales' theorem can be phrased differently too:
THEOREM: If the centre of the circumscribed circle of a triangle is the mid-point of one of the sides, then the triangle is a right-angled triangle, and the centre of the circumscribed circle is the mid-point of the hypotenuse.
Later on we are going to prove that the converse of the theorem is also true:
THEOREM: The centre of the circumscribed circle of a rightangled triangle is the mid-point of the hypotenuse.
The theorem and its converse together imply that:
THEOREM: The vertex at the right angle in every right-angled triangle drawn above a line segment as the hypotenuse lies on the circle drawn with the line
segment as the diameter. This circle is called the Thales' circle of the line segment.
Example 1
Let us construct a tangent line through a given external point to a given circle.
Solution
Since the radius drawn to the point of tangency is perpendicular to the tangent line, from Thales' theorem the circle drawn with the line segment,connecting centre O of the circle and the given external point P, as thediameter intersects the given circle at the point of tangency. The Thales'circle drawn with line segment OP as the diameter intersects the given circle at two points, therefore we get two suitable tangent lines. (Figure 78)
The steps of the construction are as follows: (Figure 79)
\latex{E_1}
\latex{E_2}
Figure 78
\latex{ O }
\latex{ F }
\latex{ P }
\latex{E_1}
\latex{E_2}
Figure 79
\latex{E_1}
\latex{E_2}
\latex{ O }
\latex{ F }
\latex{ P }
\latex{ O }
\latex{ F }
\latex{ P }
\latex{ F }
\latex{ P }
\latex{ O }
- Constructing mid-point F of line segment OP.
- Drawing the circle with centre F and with radius OF = FP. The intersection points of the two circles are \latex{E_1} and \latex{E_2}.
- Drawing straight lines \latex{PE_1} and \latex{PE_2}.
Line segments \latex{PE_1} and \latex{PE_2} are called tangent line segments.
According to the solution \latex{PE_1 = PE_2}, with this we have shown the
following theorem:
following theorem:
THEOREM: The tangent line segments drawn from an external point to the circle are of equal length.
Example 2
Let us prove that the two end-points of one side of the triangle and the base points of the altitudes starting from these points lie on one circle.
Solution
Remember that the base point of the altitude of the triangle is the intersection point of the straight line containing the altitude and the straight line of the corresponding side.
Triangles \latex{ABT_a} and \latex{ABT_b} are right-angled triangles with a common hypotenuse: side AB of the triangle (Figure 80). Resulting from the converse of the Thales' theorem the circumscribed circles of these two right-angled triangles are equivalent circles, namely the Thales' circle drawn with side AB as the diameter.
So the statement of the example is actually the consequence of the converse of the Thales' theorem.
\latex{T_{a}}
\latex{T_{b}}
\latex{m_{b}}
\latex{m_{a}}
Figure 80
\latex{ A }
\latex{ B }
\latex{ c }
\latex{ b }
\latex{ a }
\latex{ C }
\latex{E_3}
\latex{E_1}
\latex{E_2}
Figure 81
\latex{ C }
\latex{ B }
\latex{ A }
\latex{ x }
\latex{ x }
\latex{ c }
\latex{ b }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ y }
\latex{ a }
\latex{ O }
\latex{ y }
Example 3
Let us prove that the length of the diameter of the inscribed circle of the right-angled triangle is less than the sum of the length of the two legs by the length of the hypotenuse.
Solution
While solving example 1 we proved that the tangent line segments drawn from an external point to the circle are of equal length. Considering this and using the notations of figure 81
\latex{CE_1=CE_2=r}; \latex{\;E_2A=AE_3=x}; \latex{\;E_3B=BE_1=y}.
The sum of the length of the two legs:
\latex{a+b=x+y+2r} (1)
The length of the hypotenuse:
\latex{c=x+y}. (2)
When substituting (2) into (1) we get
\latex{a+b=c+2r},
and this is what we wanted to prove.
Exercises
{{exercise_number}}. The radius of the circumscribed circle of a right-angled triangle is 5 cm, the length of one of its legs is
- \latex{6 \;cm;}
- \latex{8 \;cm;}
- \latex{2 \;cm;}
- \latex{a}.
Calculate the length of the other leg and of the hypotenuse.
{{exercise_number}}. The distance of centre O of a circle with 4 cm long radius and of point P in the plane is
- \latex{5\;cm;}
- \latex{7\;cm;}
- \latex{12\;cm;}
- \latex{23 \;cm.}
Calculate the length of the tangent line segment that can be drawn from point \latex{P} to the circle in each case.
{{exercise_number}}. Prove that the base points of the two altitudes of the triangle are equidistant from the mid-point of the third side.
{{exercise_number}}. Construct a triangle if the base points of two altitudes and the straight line containing the third side are given. Investigate the number of solutions depending on the mutual position of the points and the straight line.
{{exercise_number}}. Take an isosceles triangle. Construct a circle with one of the legs of the triangle as the diameter. Where does this circle intersect the base of the triangle? (Justify your answer.)
{{exercise_number}}. The lengths of the hypotenuse and of the altitude belonging to the hypotenuse of a right-angled triangle are given. Construct the triangle. Examine the conditions of solvability.
{{exercise_number}}. Show that the circles constructed with two sides of an acute triangle as the diameters intersect each other on the third side.
{{exercise_number}}. The length of the hypotenuse and one of the legs of a right-angled triangle are
- \latex{5\;cm,\; 3\;cm;}
- \latex{13\;cm,\; 5\;cm;}
- \latex{10\;cm,\; 8\;cm;}
- \latex{28\;cm,\; 11\;cm.}
Calculate the length of the other leg and of the radius of the inscribed circle in each case.
Puzzle
In the figure \latex{EC = 6}. Give the perimeter of triangle ABC.
\latex{ A }
\latex{ B }
\latex{ E }
\latex{ C }
\latex{ D }