A kosarad üres
Applications of point reflection
The proof of the converse of the Thales’ theorem
THEOREM: The centre of the circumscribed circle of a right-angled triangle is the midpoint of the hypotenuse.
Proof
Let us reflect right-angled triangle ABC in midpoint F of hypotenuse AB. (Figure 34)
The union of the original triangle and the image triangle is a parallelogram with \latex{ 90º } angles, i.e. it is a rectangle. The rectangle is both centrally symmetric and axially symmetric which implies that its diagonals bisect each other and are of equal length. Therefore FA = FB = FC = FC’, i.e. the centre of the circumscribed circle of the rectangle is F. This circle is also the circumscribed circle of rightangled triangle ABC, thus we have shown the converse of the Thales' theorem.
Figure 34
\latex{\alpha}
\latex{\beta}
\latex{\beta}
\latex{\alpha}
\latex{\alpha+\beta=90°}
\latex{ A }
\latex{ B }
\latex{ C }
\latex{ C' }
\latex{ F }
The midline of a parallelogram, of a triangle and of a trapezium
DEFINITION: The line segment connecting the midpoints of two opposite sides of a quadrilateral is the midline of the quadrilateral.
Every quadrilateral has two midlines. By using the notation of figure 35 line segments F1F3 and F2F4 are the midlines of the quadrilateral.
Figure 35
\latex{F_{1} }
\latex{F_{2} }
\latex{F_{3} }
\latex{F_{4} }
\latex{ A }
\latex{ B }
\latex{ C }
\latex{ D }
Figure 36
\latex{FF'=CD=AB} and \latex{FF'\left\|CD\right\| AB}
\latex{ F }
\latex{ F }
\latex{ D }
\latex{ C }
\latex{ A }
\latex{ B }
Let us take a parallelogram and let us draw one of its midlines (Figure 36). In quadrilateral FF’CD FD = F’C and FD are parallel with F’C which implies that quadrilateral FF’CD is a parallelogram. Thus FF’ = CD and FF’ are parallel with CD. We can make a similar remark if we examine the other midline of the parallelogram.
Thus we have proven the following:
THEOREM: The midline of a parallelogram is parallel and of equal length with two sides of the parallelogram.
⯁ ⯁ ⯁
DEFINITION: The line segment connecting the midpoints of two sides of a triangle is a midline of the triangle. (Figure 37)
Figure 37
\latex{ F_{b} }
\latex{ F_{a} }
\latex{ F_{c} }
\latex{ C }
\latex{ B }
\latex{ A }
Every triangle has three midlines.
Let us consider triangle ABC, and let us reflect it in the midpoint of its side BC (Figure 38). When uniting the original triangle and the image triangle we get a centrally symmetric quadrilateral, parallelogram ABA’C.
The image of midpoint\latex{ F_{b} } of side AC is midpoint \latex{ F_{b} } ’ of side BA’.
So in parallelogram ABA’C line segment \latex{ F_{b} F_{b} } ’ is a midline, therefore FbFb’ is parallel with AB, and FbFb’ = AB. Because of the reflection Fa bisects line segment FbFb’, i.e. \latex{F_{a} F_{b} =\frac{AB}{2}} and FaFb are parallel with AB.
With the help of the reflection in the midpoints of the other two sides of the triangle a similar property is resulting regarding the other two midlines too.
Figure 38
\latex{ F_{a} }
\latex{ F_{b} }
\latex{ F'_{b} }
\latex{F_{a} F_{b} =\frac{AB}{2} =\frac{c}{2} } and \latex{F_{a} F_{b}\parallel AB}
\latex{ A }
\latex{ B }
\latex{ C }
\latex{ A' }
\latex{ c }
Thus we have proven the following theorem:
THEOREM: The line segment connecting the midpoints of two sides of the triangle (the midline of the triangle) is parallel with the third side of the triangle and is half the length of the third side of the triangle.
⯁ ⯁ ⯁
Let us take trapezium ABCD, and let us reflect it in midpoint Fb of leg BC (Figure 39). The union of the original trapezium and the image trapezium is parallelogram AD’A’D.
Figure 39
\latex{ F_{d} }
\latex{ F_{b} }
\latex{ F'_{d} }
\latex{F_{b} F_{d}=\frac{a+c}{2} }
\latex{F_{b} F_{d}\parallel AB\parallel CD}
\latex{F_{b} F_{d}\parallel AB\parallel CD}
\latex{ A }
\latex{ B }
\latex{ C }
\latex{ D }
\latex{ a }
\latex{ c }
\latex{ b }
\latex{ d }
\latex{ a }
\latex{ c }
\latex{ D' }
\latex{ A' }
The image of midpoint Fd of side AD is midpoint Fd’ of side A’D’. So in parallelogram AD’A’D line segment FdFd’ is a midline, therefore
\latex{F_{d} F'_{d}=AD=AB+BD'=AB+CD=a+c}
Because of the reflection Fb bisects line segment FdFd’, i.e.
\latex{F_{d} F'_{d}=\frac{AB+CD}{2} =\frac{a+c}{2}}
and FbFd is parallel with AB and CD. Thus we have justified the following:
THEOREM: The midline connecting the midpoints of the legs of a trapezium is parallel with the bases of the trapezium, and its length is half of the sum of the length of the bases (the arithmetic mean of the length of the two bases).
The (straight lines containing the) altitudes of the triangle
DEFINITION: The altitude of a triangle is the perpendicular line segment from a vertex to the straight line of the opposite side. (Figure 40)
Figure 40
\latex{ h_{b} }
\latex{ h_{a} }
\latex{ h_{a} }
acute triangle
right-angled triangle
obtuse triangle
\latex{ A }
\latex{ B }
\latex{ A }
\latex{ A }
\latex{ a }
\latex{ a }
\latex{ a }
\latex{ b }
THEOREM: The altitudes of the triangle intersect each other at one point.
Figure41
\latex{ h_{b} }
\latex{ h_{c} }
\latex{ B' }
\latex{ C' }
\latex{ B }
\latex{ C }
\latex{ A' }
Proof
Let us take the triangle \latex{ ABC }, and through all three vertices let us draw a straight line parallel with the opposite side (Figure 41). Let the intersection points of two and two such straight lines be \latex{ A’, B’, C’ } respectively as in Figure 41.
Quadrilateral \latex{ ABCB’ } is a parallelogram, since its opposite sides are parallel.
Thus \latex{ AB = B’C }. For the same reason quadrilateral \latex{ ABA’C } is also a parallelogram, therefore \latex{ AB = CA’ }. Therefore \latex{ B’C = CA’ } results which means that in triangle \latex{ A’B’C’ } point \latex{ C } is the midpoint of side \latex{ A’B’ }.
Figure 42
\latex{ A }
\latex{ B }
\latex{ C }
\latex{ H }
\latex{ B }
\latex{ A }
\latex{ C=H }
\latex{ A }
\latex{ C }
\latex{ B }
\latex{ H }
By similar reasoning \latex{ A } bisects side \latex{ B’C’ }, and \latex{ B } bisects side \latex{ C’A’ }. It all means that triangle \latex{ ABC } is the triangle defined by the midlines of triangle \latex{ A’B’C’ }. The midlines are parallel with the corresponding third sides, therefore the altitudes of triangle\latex{ ABC } are perpendicular to the sides of triangle\latex{ A’B’C’ }, and they intersect them at their midpoints. Therefore the altitudes of triangle \latex{ ABC } are the perpendicular bisectors of the sides of triangle \latex{ A’B’C’ }, and about these we know from our earlier studies that they intersect each other at one point.
With this we have shown the theorem about the altitudes of the triangle.
DEFINITION: The common point of the altitudes is called the orthocentre of the triangle.
Note: When constructing the orthocentre of an acute triangle, a right-angled triangle and an obtuse triangle it can be seen that the orthocentre of an acute triangle is inside the triangle, the orthocentre of a right-angled triangle is at the right-angled vertex, and the orthocentre of an obtuse triangle is outside the triangle. (Figure 42)
The medians of a triangle
DEFINITION: The median of a triangle is the line segment connecting the vertex with the midpoint of the side opposite. (Figure 43)
THEOREM: Any two medians of the triangle intersect each other so that the intersection point divides both medians into two parts in the ratio of \latex{ 1:2 }; the other end-point of the longer part is the corresponding vertex of the triangle.
Figure 43
\latex{ s_{b} }
\latex{ s_{a} }
\latex{ F_{b} }
\latex{ F_{a} }
\latex{ C }
\latex{ A }
\latex{ B }
Proof
Let \latex{ S } denote the intersection point of medians \latex{ s_a } and \latex{ s_b } of triangle \latex{ ABC }. And let the midpoint of line segment \latex{ AS } be\latex{ P }, and let the midpoint of line segment \latex{ BS } be \latex{ Q }.
(Figure 44)
In triangle \latex{ ABC F_a \, F_b }is a midline, therefore
\latex{F_{a} F_{b}=\frac{AB}{2}},
and \latex{ F_a \, F_b } is parallel with \latex{ AB }.
Figure 44
\latex{ s_{a} }
\latex{ s_{b} }
\latex{ F_{b} }
\latex{ F_{a} }
\latex{ C }
\latex{ A }
\latex{ B }
In triangle \latex{ ABS } \latex{ PQ } is a midline, therefore
\latex{PQ=\frac{AB}{2}},
and \latex{ PQ } is parallel with \latex{ AB }.
When comparing the above two remarks we get that
FaFb = PQ,
and \latex{ F_a\,F_b } is parallel with \latex{ PQ }, which means that two opposite sides of quadrilateral \latex{ PQ }\latex{ F_a\,F_b } are parallel and of equal length. It implies that quadrilateral \latex{ PQ }\latex{ F_a\,F_b } is a parallelogram.
The diagonals of the parallelogram bisect each other, and hence
PS = SFa and QS = SFb.
Since \latex{ P } and \latex{ Q } are the midpoints of line segments \latex{ AS } and \latex{ BS } respectively, therefore
\latex{AS=2\times SF_{a}} and \latex{BS=2\times SF_{b}}.
With this we have proven the theorem.
Since any two medians trisect each other according to the description in the theorem, therefore the trisection point of the medians further away from the vertex of the triangle is a common point; all three medians pass through it. Thus the following is resulting as the consequence of the previous theorem:
Feuerbach circle
Euler line
\latex{ S }
\latex{ O }
\latex{ M }
THEOREM: The medians of the triangle intersect each other at one point. This point is the trisection point of all three medians further away from the corresponding vertex of the triangle.
DEFINITION: The intersection point of the medians is called the centroid of the triangle.
Exercises
{{exercise_number}}. The length of the sides of a triangle are:
- \latex{ 3 } \latex{ cm }; \latex{ 4 } \latex{ cm }; \latex{ 5 } \latex{ cm };
- \latex{ 6 } \latex{ dm }; \latex{ 7 } \latex{ dm }; \latex{ 10 } \latex{ dm };
- \latex{ 7.2 } \latex{ m }; \latex{ 410 } \latex{ cm }; \latex{ 50 } \latex{ dm };
- \latex{ 12 } \latex{ cm }; \latex{ 7.2 } \latex{ cm }; \latex{ 48 } \latex{ mm }.
Calculate the length of the sides of the new triangle defined by the midlines of the triangle in each case.
{{exercise_number}}. The length of the parallel sides of a trapezium are:
- \latex{ 4 } \latex{ cm } and \latex{ 8 } \latex{ cm };
- \latex{ 5 } \latex{ dm } and \latex{ 17 } \latex{ dm };
- \latex{ 12.5 } \latex{ cm } and \latex{ 0.3 } \latex{ m };
- \latex{ 32 } \latex{ mm } and \latex{ 0.62 } \latex{ dm }.
Calculate the length of the line segment connecting the midpoints of the legs in each case.
{{exercise_number}}. Construct a right-angled triangle if the midline parallel with the hypotenuse is \latex{ 3 } \latex{ cm } long and the altitude belonging to the hypotenuse is \latex{ 2 } \latex{ cm } long.
{{exercise_number}}. Calculate the radius of the circumscribed circle of a right-angled triangle if the length of the two legs are:
- \latex{ 3 } \latex{ cm } and \latex{ 4 } \latex{ cm };
- \latex{ 5 } \latex{ dm } and \latex{ 12 } \latex{ dm };
- \latex{ 12 } \latex{ mm } and \latex{ 3.5 } \latex{ cm };
- \latex{ a } and \latex{ b }.
{{exercise_number}}. Calculate the length of the altitude of a regular triangle if its centroid is the following distance away from its vertices:
- \latex{ 4 } \latex{ cm };
- \latex{ 6 } \latex{ dm };
- \latex{ 12.3 } \latex{ m };
- \latex{ d }.
{{exercise_number}}. Reflect triangle \latex{ ABC } in midpoint \latex{ Fa } of side \latex{ BC }. What planar figure is defined by the union of the original triangle and the image triangle? Calculate the length of the diagonals of the resulting planar figure if
- \latex{AF_{a}}\latex{ = 5 } \latex{ cm }; \latex{BC=}\latex{ 8 } \latex{ cm };
- \latex{AF_{a}=} \latex{ 6.2 } \latex{ dm }; \latex{BC=} \latex{ 410 } \latex{ mm };
- \latex{AF_{a}=x; BC=y}.
{{exercise_number}}. Construct a triangle if the length of two sides and the length of the median starting from the common vertex are given.
- \latex{a =}\latex{ 4.2 } \latex{ cm }; \latex{b =}\latex{ 9.2 } \latex{ cm }; \latex{s_{c} =}\latex{ 6 } \latex{ cm };
- \latex{a =}\latex{ 50 } \latex{ mm }; \latex{b =}\latex{ 0.7 } \latex{ dm }; \latex{s_{c} =}\latex{ 6 } \latex{ cm };
- \latex{a =}\latex{ 4 } \latex{ cm }; \latex{b =}\latex{ 0.73 } \latex{ dm }; \latex{s_{c} =}\latex{ 0.06 } \latex{ m }.
{{exercise_number}}. Prove that the following statement is true for any quadrilateral: the line segment connecting the midpoints of two adjacent sides is half the length of one of the diagonals of the quadrilateral.
{{exercise_number}}. Verify that the midpoints of the sides of any quadrilateral define a parallelogram.
{{exercise_number}}. Prove that the midlines of any quadrilateral bisect each other.
{{exercise_number}}. What can we state about the quadrilateral the midlines of which are of equal length? Justify your statement.
{{exercise_number}}. Prove that the mutually taken common secant lines of the circles constructed above the sides of the triangle as diameter intersect each other at one point. Which remarkable point of the triangle is this point?
{{exercise_number}}. Show that the sum of the length of the medians of the triangle is less than the perimeter of the triangle, but it is greater than \latex{ 75 }% of the perimeter.
{{exercise_number}}. Check the theorems relating to the Euler line and the Feuerbach circle with the help of a computer based geometric construction program.
