Middle line. Midlines of triangles and quadrangles
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Educational research work
Middle lines geometric shapes
Morozova Elizabeth
Gomel 2010
Introduction
1.The properties of the middle lines
2. Triangle, quadrilateral, parallelogram
3. Quadrangle, tetrahedron. Centers of mass
4. Tetrahedron, octahedron, parallelepiped, cube
Conclusion
List of used literature
Application
Introduction
Geometry is an integral part of the general culture, and geometric methods serve as a tool for understanding the world, contribute to the formation of scientific ideas about the surrounding space, the disclosure of the harmony and perfection of the Universe. Geometry starts with a triangle. For two millennia, the triangle has been a symbol of geometry, but it is not a symbol. The triangle is an atom of geometry. The triangle is inexhaustible - its new properties are constantly being discovered. To tell about all its known properties, a volume is needed that is comparable in volume to the volume of the Great Encyclopedia. We want to talk about the middle lines of geometric shapes and their properties.
In our work, a chain of theorems is traced that covers the entire course of geometry. It starts with the triangle midline theorem and leads to interesting properties of the tetrahedron and other polyhedra.
The middle line of the shapes is the segment connecting the midpoints of the two sides of the given shape.
1. Properties of the middle lines
Triangle properties:
when all three middle lines are drawn, 4 equal triangles are formed, similar to the original one with a coefficient of 1/2.
the middle line is parallel to the base of the triangle and equal to half of it;
the middle line cuts off a triangle that is similar to this one, and its area is equal to one quarter of its area.
Quadrilateral properties:
if in a convex quadrilateral the middle line makes equal angles with the diagonals of the quadrilateral, then the diagonals are equal.
the length of the center line of the quadrilateral is less than half the sum of the other two sides or equal to it if these sides are parallel, and only in this case.
the midpoints of the sides of an arbitrary quadrilateral are the vertices of the parallelogram. Its area is half the area of a quadrilateral, and its center lies at the intersection of the midlines. This parallelogram is called the Varignon parallelogram;
The point of intersection of the midlines of the quadrilateral is their common midpoint and halves the segment connecting the midpoints of the diagonals. In addition, it is the centroid of the vertices of the quadrilateral.
Trapezium properties:
the middle line is parallel to the bases of the trapezoid and is equal to their half-sum;
the midpoints of the sides of an isosceles trapezoid are the vertices of the rhombus.
2. Triangle, quadrilateral, parallelogram
Three equal triangles AKM, BLK, CLM can be attached to any triangle KLM, each of which together with the triangle KLM forms a parallelogram (Fig. 1). In this case, AK = ML = KB, and three angles equal to three different angles triangle, totaling 180 °, therefore K is the middle of the segment AB; similarly, L is the midpoint of segment BC, and M is the midpoint of segment CA.
Theorem 1... If we connect the midpoints of the sides in any triangle, we get four equal triangles, with the middle being a parallelogram with each of the other three.
In this formulation, all three middle lines of the triangle participate at once.
Theorem 2... The segment connecting the midpoints of the two sides of the triangle is parallel to the third side of the triangle and is equal to its half (see Fig. 1).
It is this theorem and its inverse - that a straight line parallel to the base and passing through the middle of one lateral side of a triangle, halves the other lateral side as well - are most often needed when solving problems.
The theorem on the midlines of a triangle implies the property of the midline of a trapezoid (Fig. 2), as well as the theorem on segments connecting the midpoints of the sides of an arbitrary quadrilateral.
Theorem 3... The midpoints of the sides of the quadrilateral are the vertices of the parallelogram. The sides of this parallelogram are parallel to the diagonals of the quadrilateral, and their lengths are equal to half the lengths of the diagonals.
Indeed, if K and L are the midpoints of sides AB and BC (Fig. 3), then KL is middle line triangle ABC, therefore the segment KL is parallel to the diagonal AC and is equal to its half; if M and N are the midpoints of sides CD and AD, then the segment MN is also parallel to AC and equal to AC / 2. Thus, the segments KL and MN are parallel and equal to each other, which means that the quadrilateral KLMN is a parallelogram.
As a consequence of Theorem 3, we obtain interesting fact(vol. 4).
Theorem 4... In any quadrilateral, the line segments connecting the midpoints of opposite sides are halved by the intersection point.
In these segments you can see the diagonals of the parallelogram (see Fig. 3), and in the parallelogram the diagonals are divided by the point of intersection in half (this point is the center of symmetry of the parallelogram).
We see that Theorems 3 and 4 and our reasoning remain true for both a non-convex quadrangle and a self-intersecting quadrangular closed polyline (Fig. 4; in the latter case, it may turn out that the parallelogram KLMN is "degenerate" - points K, L, M, N lie on one straight line).
Let us show how we can derive the main theorem on the medians of a triangle from Theorems 3 and 4.
Theorem5 ... The medians of the triangle intersect at one point and divide it in a ratio of 2: 1 (counting from the vertex from which the median is drawn).
Let's draw two medians AL and SC of triangle ABC. Let O be the point of their intersection. The midpoints of the nonconvex quadrilateral ABCO - points K, L, M and N (Fig. 5) - are the vertices of the parallelogram, and the point of intersection of its diagonals KM and LN for our configuration will be the point of intersection of the medians O. So, AN = NO = OL and CM = MO = OK, i.e. point O divides each of the medians AL and CK in a 2: 1 ratio.
Instead of the median CK, we could consider the median drawn from the vertex B, and make sure in the same way that it divides the median AL in the ratio 2: 1, i.e., it passes through the same point O.
3. Quadrangle and tetrahedron. Centers of mass
Theorems 3 and 4 are also true for any spatial closed polyline of four links AB, BC, CD, DA, four vertices A, B, C, D of which do not lie in the same plane.
Such a spatial quadrangle can be obtained by cutting out a quadrilateral ABCD from paper and bending it diagonally at a certain angle (Fig. 6, a). It is clear that the middle lines KL and MN of triangles ABC and ADC remain as before their middle lines and will be parallel to the segment AC and equal to AC / 2. (Here we use the fact that the main property of parallel lines remains true for space: if two lines KL and MN are parallel to the third line AC, then KL and MN lie in the same plane and are parallel to each other.)
Thus, points K, L, M, N are the tops of the parallelogram; thus, the segments KM and LN intersect and are halved by the intersection point. Instead of a quadrangle, here we can talk about a tetrahedron - a triangular pyramid ABCD: the midpoints K, L, M, N of its edges AB, AC, CD and DA always lie in the same plane. Cutting the tetrahedron along this plane (Fig. 6, b), we get a parallelogram KLMN, two sides of which are parallel to the edge AC and equal
AC / 2, and the other two are parallel to the edge BD and equal to BD / 2.
The same parallelogram - the "middle section" of the tetrahedron - can be constructed for other pairs of opposite edges. Each two of these three parallelograms have a common diagonal. In this case, the midpoints of the diagonals coincide. So, we get an interesting consequence:
Theorem 6... Three line segments connecting the midpoints of opposite edges of the tetrahedron intersect at one point and are divided by it in half (Fig. 7).
This and other facts discussed above are naturally explained in the language of mechanics - with the help of the concept of the center of mass. Theorem 5 speaks of one of the remarkable points of the triangle — the point of intersection of the medians; in Theorem 6 - o wonderful point for the four vertices of the tetrahedron. These points are the centers of mass of the triangle and tetrahedron, respectively. Let us first return to Theorem 5 on the medians.
We place three identical weights at the vertices of the triangle (Fig. 8).
Let us take the mass of each as a unit. Let's find the center of mass of this system of weights.
Let us first consider two weights located at the vertices A and B: their center of mass is located in the middle of the segment AB, so that these weights can be replaced with one weight of mass 2, placed in the middle K of the segment AB (Fig. 8, a). Now you need to find the center of mass of a system of two weights: one with mass 1 at point C and the second with mass 2 at point K. According to the lever rule, the center of mass of such a system is at point O, dividing the segment SK in a ratio of 2: 1 (closer to the load at point K with greater mass - Fig. 8, b).
We could first combine the loads at points B and C, and then - the resulting load of mass 2 in the middle L of the segment BC - with the load at point A. Or, first, combine loads A and C, and. then attach B. In any case, we should get the same result. The center of mass is, therefore, at point O, dividing each of the medians in a ratio of 2: 1, counting from the top. Theorem 4 could be explained by similar considerations - the fact that the segments connecting the midpoints of opposite sides of the quadrilateral divide each other in half (serve as the diagonals of the parallelogram): it is enough to place identical weights at the vertices of the quadrilateral and combine them in pairs in two ways (Fig. 9).
Of course, four unit weights located on a plane or in space (at the vertices of a tetrahedron) can be split into two pairs in three ways; the center of mass is in the middle between the midpoints of the segments connecting these pairs of points (Fig. 10) - explanation of Theorem 6. (For a flat quadrangle, the result looks like this: two segments connecting the midpoints of opposite sides and a segment connecting the midpoints of the diagonals intersect at one point Oh and share it in half).
Through point O - the center of mass of four identical weights - four more segments pass, connecting each of them with the center of mass of the other three. These four segments are divided by point O in a ratio of 3: 1. To explain this fact, you must first find the center of mass of three weights and then attach the fourth.
4. Tetrahedron, octahedron, parallelepiped, cube
At the beginning of the work, we considered a triangle divided by the middle lines into four identical triangles (see Fig. 1). Let's try to do the same construction for an arbitrary triangular pyramid (tetrahedron). We cut the tetrahedron into parts as follows: through the midpoints of three edges emerging from each vertex, we make a flat cut (Fig. 11, a). Then four identical small tetrahedrons will be cut off from the tetrahedron. By analogy with the triangle, one would think that one more tetrahedron of the same type would remain in the middle. But this is not so: the polyhedron, which remains from the large tetrahedron after removing four small ones, will have six vertices and eight faces - it is called an octahedron (Fig. 11.6). It is convenient to test this using a tetrahedral piece of cheese. The resulting octahedron has a center of symmetry, since the midpoints of the opposite edges of the tetrahedron intersect at common point and divide it in half.
One interesting construction is connected with the triangle, divided by the middle lines into four triangles: we can consider this figure as a development of some tetrahedron.
Imagine an acute-angled triangle cut out of paper. Having bent it along the middle lines so that the vertices converge at one point, and gluing the edges of the paper converging at this point, we get a tetrahedron, in which all four faces are equal triangles; its opposite edges are equal (fig. 12). Such a tetrahedron is called semiregular. Each of the three "middle sections" of this tetrahedron - parallelograms whose sides are parallel to opposite edges and equal to their halves - will be a rhombus.
Therefore, the diagonals of these parallelograms - three line segments connecting the midpoints of opposite edges - are perpendicular to each other. Among the numerous properties of a semi-regular tetrahedron, we note the following: the sum of the angles converging at each of its vertices is 180 ° (these angles are respectively equal to the angles of the original triangle). In particular, if we start with a sweep in the shape of an equilateral triangle, we get a regular tetrahedron, in which
At the beginning of the work, we saw that each triangle can be viewed as a triangle formed by the midlines of the larger triangle. There is no direct analogy in space for such a construction. But it turns out that any tetrahedron can be regarded as the "core" of a parallelepiped, in which all six edges of the tetrahedron serve as diagonals of the faces. To do this, you need to do the following construction in space. Draw a plane parallel to the opposite edge through each edge of the tetrahedron. The planes drawn through the opposite edges of the tetrahedron will be parallel to each other (they are parallel to the plane of the "middle section" - a parallelogram with vertices at the midpoints of the other four edges of the tetrahedron). Thus, three pairs of parallel planes are obtained, at the intersection of which the desired parallelepiped is formed (two parallel planes intersect the third along parallel lines). The vertices of the tetrahedron serve as four non-adjacent vertices of the constructed parallelepiped (Fig. 13). Conversely, in any parallelepiped, you can select four non-adjacent vertices and cut off corner tetrahedrons from it by planes passing through every three of them. After that, there will be a "core" - a tetrahedron, the edges of which are the diagonals of the parallelepiped faces.
If the original tetrahedron is semiregular, then each face of the constructed parallelepiped will be a parallelogram with equal diagonals, i.e. rectangle.
The converse is also true: the "core" of a rectangular parallelepiped is a semi-regular tetrahedron. Three rhombuses - the middle sections of such a tetrahedron - lie in three mutually perpendicular planes. They serve as planes of symmetry for an octahedron obtained from such a tetrahedron by cutting off the corners.
For a regular tetrahedron, the parallelepiped described around it will be a cube (Fig. 14), and the centers of the faces of this cube - the midpoints of the edges of the tetrahedron - will be the vertices of a regular octahedron, all the faces of which are regular triangles... (The three planes of symmetry of the octahedron intersect the tetrahedron in squares.)
Thus, in Figure 14 we see three of the five Platonic solids (regular polyhedrons) at once - a cube, a tetrahedron and an octahedron.
Conclusion
Based on the work done, the following conclusions can be drawn:
The middle lines have different beneficial features in geometric shapes.
One theorem can be proved using the middle line of figures, and also explained in the language of mechanics - using the concept of the center of mass.
Using the middle lines, you can build various planimetric (parallelogram, rhombus, square) and stereometric shapes (cube, octahedron, tetrahedron, etc.).
The properties of the middle lines help to rationally solve problems of any level.
List of sources and literature used
Monthly popular scientific physics and mathematics journal of the Academy of Sciences of the USSR and the Academy of Pedagogical Sciences of Literature. “Quantum No. 6 1989, p. 46.
S. Aksimova. Entertaining mathematics. - St. Petersburg, "Trigon", 1997, p. 526.
V.V. Shlykov, L.E. Zezetko. Practical classes in geometry, grade 10: a guide for teachers. - Minsk: TetraSystems, 2004, p. 68.76, 78.
Application
Why can't the middle line of a trapezoid go through the intersection of the diagonals?
BCDA 1 B 1 C 1 D 1 - parallelepiped. Points E and F are the intersection points of the diagonals of the faces. AA1B 1 B and BB 1 C 1 C, respectively, and points K and T are the midpoints of the edges AD and DC, respectively. Is it true that lines EF and CT are parallel?
In the triangular prism ABCA 1 B 1 C 1 points O and F are the midpoints of the ribs AB and BC, respectively. Points T and K are the middle of the segments AB 1 and BC 1, respectively. How are the direct TCs and OFs located?
ABCA 1 B 1 C 1 is a regular triangular prism, all edges of which are equal to each other. Point O is the midpoint of the edge CC 1, and point F lies on the edge BB] so that BF: FB X = 1: 3. Construct a point K at which the straight line l passing through the point F parallel to the straight line AO intersects the plane ABC. Calculate the total surface area of the prism if KF = 1 cm.
figure
Earlier. 2. This geometric figure... This figure formed closed line... They are convex and non-convex. Have figures there are sides ..., sector, sphere, segment, sine, middle, average line, ratio, property, degree, stereometry, secant ...
Midlines of quadrangles and their properties Completed by: Matveev Dmitry Teacher: Rychkova Tatyana Viktorovna Lyceum "Dubna" 9IM 2007 Average lines and Varignon Parallelogram Other properties of the midline of a quadrangle A short list of all theorems and properties
What is a Varignon parallelogram? This is a parallelogram, the vertices of which are the midpoints of the sides of the quadrilateral Otherwise: it is a parallelogram, the diagonals of which are the midlines of the quadrilateral
A B C D N M L K P Proof: Connect the points K, L, M, N and draw a diagonal AC; In ∆ACD, NM is the middle line, which means NM AC and NM = 1/2 AC; In ∆ABC, KL is the middle line, so KL AC and KL = 1/2 AC; NM = 1/2 AC = KL, NM AC KL, so the quadrangle KLMN is a parallelogram. A L B M C D K P N Proof: Connect the points K, L, M, N and draw a diagonal DB; In ∆CDB, NM is the middle line, which means NM DB and NM = 1/2 DB; In ∆ADC, KL is the middle line, so KL DB and KL = 1/2 DB; NM = 1/2 DB = KL, NM DB KL, so the quadrilateral KLMN is a parallelogram. Let us prove that KLMN is the Varignon parallelogram, with KM and NM - the middle lines ABCD.
This means ... Since the KLMN quadrilateral is a Varignon parallelogram, its diagonals at the intersection point are halved. The middle lines of any quadrilateral are halved
Consequences: 1. If the midlines of the quadrilateral are equal, then the midpoints of the sides of the quadrilateral (the vertices of the Varignon parallelogram) lie on the same circle. Proof: Since in the Varignon parallelogram equal centerlines are equal diagonals, this parallelogram is a rectangle, and around it you can always describe a circle, which means that its vertices lie on one circle.
Corollaries: 2. If the midlines of the quadrilateral are perpendicular, then the diagonals of the quadrilateral are equal. Proof: Since NL┴KM and NL with KM are diagonals in the parallelogram KLMN, then KLMN is a rhombus. Therefore, KL = LM = MN = NK. Since AC = 2 KL and BD = 2 NK, then AC = BD. A K B L C M D N P O A P K C D M N L B
Corollaries: A K B L C M D N P O A P K C D M N L B 3. If the diagonals of the quadrilateral are equal, then the midlines of the quadrilateral are perpendicular. Proof: Since AC = 2 MN = 2 KL, BD = 2 NK = 2 ML and AC = BD, then KL = LM = MN = NK. This means that KLMN is a rhombus, and the diagonals in the rhombus are perpendicular, that is, NL┴KM.
For example: Solving such a problem would have to work hard without knowing one of the properties of the Varignon parallelogram:
What is the area of the Varignon parallelogram? Proof for a convex quadrilateral: Consider ∆ABD and ∆ANK: a).
middle line figures in planimetry - a segment connecting the midpoints of two sides of a given figure. The concept is used for the following figures: triangle, quadrilateral, trapezoid.
Middle line of a triangle
Properties
- the middle line of the triangle is parallel to the base and equal to half of it.
- the middle line cuts off a triangle similar and homothetic to the original one with a factor of 1/2; its area is equal to one-fourth the area of the original triangle.
- the three center lines divide the original triangle into four equal triangles. The central of these triangles is called the complementary or median triangle.
Signs
- if a segment is parallel to one of the sides of the triangle and connects the midpoint of one side of the triangle with a point lying on the other side of the triangle, then this is the middle line.
Midline of a quadrilateral
Midline of a quadrilateral- a segment connecting the midpoints of the opposite sides of the quadrilateral.
Properties
The first line connects 2 opposite sides. The second connects 2 other opposite sides. The third connects the centers of the two diagonals (not all quadrangles have the point of intersection that divides the diagonals in half).
- If in a convex quadrilateral the middle line forms equal angles with the diagonals of the quadrilateral, then the diagonals are equal.
- The length of the center line of the quadrilateral is less than half the sum of the other two sides or equal to it if these sides are parallel, and only in this case.
- The midpoints of the sides of an arbitrary quadrilateral are the vertices of the parallelogram. Its area is half the area of a quadrilateral, and its center lies at the intersection of the midlines. This parallelogram is called the Varignon parallelogram;
- The last point means the following: In a convex quadrilateral, four middle lines of the second kind. Middle lines of the second kind- four segments inside the quadrilateral passing through the midpoints of its adjacent sides parallel to the diagonals. Four middle lines of the second kind a convex quadrilateral, cut it into four triangles and one central quadrilateral. This central quadrilateral is the Varignon parallelogram.
- The point of intersection of the midlines of the quadrilateral is their common midpoint and halves the segment connecting the midpoints of the diagonals. In addition, it is the centroid of the vertices of the quadrilateral.
- In an arbitrary quadrilateral, the midline vector is equal to the half-sum of the base vectors.
The middle line of the trapezoid
The middle line of the trapezoid
The middle line of the trapezoid- a segment connecting the midpoints of the lateral sides of this trapezoid. The segment connecting the midpoints of the bases of the trapezoid is called the second midline of the trapezoid.
It is calculated by the formula: E F = A D + B C 2 (\ displaystyle EF = (\ frac (AD + BC) (2))), where AD and BC- the base of the trapezoid.
Introduction
Geometry is an integral part of the general culture, and geometric methods serve as a tool for understanding the world, contribute to the formation of scientific ideas about the surrounding space, the disclosure of the harmony and perfection of the Universe. Geometry starts with a triangle. For two millennia, the triangle has been a symbol of geometry, but it is not a symbol. The triangle is an atom of geometry. The triangle is inexhaustible - its new properties are constantly being discovered. To tell about all its known properties, a volume is needed that is comparable in volume to the volume of the Great Encyclopedia. We want to talk about the middle lines of geometric shapes and their properties.
In our work, a chain of theorems is traced that covers the entire course of geometry. It starts with the triangle midline theorem and leads to interesting properties of the tetrahedron and other polyhedra.
The middle line of the shapes is the segment connecting the midpoints of the two sides of the given shape.
Midline properties
Triangle properties:
· When all three middle lines are drawn, 4 equal triangles are formed, similar to the original one with a coefficient of 1/2.
· The middle line is parallel to the base of the triangle and is equal to half of it;
· The middle line cuts off a triangle that is similar to this one, and its area is equal to one quarter of its area.
Quadrilateral properties:
· If in a convex quadrilateral the middle line makes equal angles with the diagonals of the quadrilateral, then the diagonals are equal.
· The length of the midline of the quadrilateral is less than half the sum of the other two sides or equal to it if these sides are parallel, and only in this case.
· The midpoints of the sides of an arbitrary quadrilateral - the vertices of the parallelogram. Its area is half the area of a quadrilateral, and its center lies at the intersection of the midlines. This parallelogram is called the Varignon parallelogram;
· The point of intersection of the midlines of the quadrilateral is their common midpoint and halves the segment connecting the midpoints of the diagonals. In addition, it is the centroid of the vertices of the quadrilateral.
Trapezium properties:
· The middle line is parallel to the bases of the trapezoid and is equal to their half-sum;
· The midpoints of the sides of an isosceles trapezoid are the vertices of a rhombus.
Triangle, quadrangle, parallelogram
Three equal triangles AKM, BLK, CLM can be attached to any triangle KLM, each of which together with the triangle KLM forms a parallelogram (Fig. 1). In this case, AK = ML = KB, and three angles, equal to three different angles of the triangle, are adjacent to the vertex of the K, making up 180 ° in total, therefore K is the middle of the segment AB; similarly, L is the midpoint of segment BC, and M is the midpoint of segment CA.
Theorem 1... If we connect the midpoints of the sides in any triangle, we get four equal triangles, with the middle being a parallelogram with each of the other three.
In this formulation, all three middle lines of the triangle participate at once.
Theorem 2... The segment connecting the midpoints of the two sides of the triangle is parallel to the third side of the triangle and is equal to its half (see Fig. 1).
It is this theorem and its inverse - that a straight line parallel to the base and passing through the middle of one lateral side of a triangle, halves the other lateral side as well - are most often needed when solving problems.
The theorem on the midlines of a triangle implies the property of the midline of a trapezoid (Fig. 2), as well as the theorem on segments connecting the midpoints of the sides of an arbitrary quadrilateral.
Theorem 3... The midpoints of the sides of the quadrilateral are the vertices of the parallelogram. The sides of this parallelogram are parallel to the diagonals of the quadrilateral, and their lengths are equal to half the lengths of the diagonals.
Indeed, if K and L are the midpoints of sides AB and BC (Fig. 3), then KL is the middle line of triangle ABC, therefore the segment KL is parallel to the diagonal AC and is equal to its half; if M and N are the midpoints of sides CD and AD, then the segment MN is also parallel to AC and equal to AC / 2. Thus, the segments KL and MN are parallel and equal to each other, which means that the quadrilateral KLMN is a parallelogram.
As a corollary to Theorem 3, we obtain an interesting fact (point 4).
Theorem 4... In any quadrilateral, the line segments connecting the midpoints of opposite sides are halved by the intersection point.
In these segments you can see the diagonals of the parallelogram (see Fig. 3), and in the parallelogram the diagonals are divided by the point of intersection in half (this point is the center of symmetry of the parallelogram).
We see that Theorems 3 and 4 and our reasoning remain true for both a non-convex quadrangle and a self-intersecting quadrangular closed polyline (Fig. 4; in the latter case, it may turn out that the parallelogram KLMN is "degenerate" - points K, L, M, N are collinear).
Let us show how we can derive the main theorem on the medians of a triangle from Theorems 3 and 4.
Theorem 5... The medians of the triangle intersect at one point and divide it in a ratio of 2: 1 (counting from the vertex from which the median is drawn).
Let's draw two medians AL and SC of triangle ABC. Let O be the point of their intersection. The midpoints of the nonconvex quadrilateral ABCO - points K, L, M and N (Fig. 5) - are the vertices of the parallelogram, and the point of intersection of its diagonals KM and LN for our configuration will be the point of intersection of the medians O. So, AN = NO = OL and CM = MO = OK, i.e. point O divides each of the medians AL and SK in a 2: 1 ratio.
Instead of the median CK, we could consider the median drawn from the vertex B, and make sure in the same way that it divides the median AL in the ratio 2: 1, i.e., it passes through the same point O.
3. Quadrangle and tetrahedron. Centers of mass
Theorems 3 and 4 are also true for any spatial closed polyline of four links AB, BC, CD, DA, four vertices A, B, C, D of which do not lie in the same plane.
Such a spatial quadrangle can be obtained by cutting out a quadrilateral ABCD from paper and bending it diagonally at a certain angle (Fig. 6, a). It is clear that the middle lines KL and MN of triangles ABC and ADC remain as before their middle lines and will be parallel to the segment AC and equal to AC / 2. (Here we use the fact that the main property of parallel lines remains true for space: if two lines KL and MN are parallel to the third line AC, then KL and MN lie in the same plane and are parallel to each other.)
Thus, points K, L, M, N are the tops of the parallelogram; thus, the segments KM and LN intersect and are halved by the intersection point. Instead of a quadrangle, here we can talk about a tetrahedron - a triangular pyramid ABCD: the midpoints K, L, M, N of its edges AB, AC, CD and DA always lie in the same plane. Cutting the tetrahedron along this plane (Fig. 6, b), we get a parallelogram KLMN, two sides of which are parallel to the edge AC and equal
AC / 2, and the other two are parallel to the edge BD and equal to BD / 2.
The same parallelogram - the "middle section" of the tetrahedron - can be constructed for other pairs of opposite edges. Each two of these three parallelograms have a common diagonal. In this case, the midpoints of the diagonals coincide. So, we get an interesting consequence:
Theorem 6... Three line segments connecting the midpoints of opposite edges of the tetrahedron intersect at one point and are divided by it in half (Fig. 7).
This and other facts discussed above are naturally explained in the language of mechanics - with the help of the concept of the center of mass. Theorem 5 speaks of one of the remarkable points of the triangle — the point of intersection of the medians; in Theorem 6 - about a remarkable point for the four vertices of the tetrahedron. These points are the centers of mass of the triangle and tetrahedron, respectively. Let us first return to Theorem 5 on the medians.
We place three identical weights at the vertices of the triangle (Fig. 8).
Let us take the mass of each as a unit. Let's find the center of mass of this system of weights.
Let us first consider two weights located at the vertices A and B: their center of mass is located in the middle of the segment AB, so that these weights can be replaced with one weight of mass 2, placed in the middle K of the segment AB (Fig. 8, a). Now you need to find the center of mass of a system of two weights: one with mass 1 at point C and the other with mass 2 at point K. According to the lever rule, the center of mass of such a system is located at point O, dividing the segment SK in a ratio of 2: 1 (closer to load at point K with greater mass - Fig. 8, b).
We could first combine the loads at points B and C, and then - the resulting load of mass 2 in the middle L of the segment BC - with the load at point A. Or, first, combine loads A and C, and. then attach B. In any case, we should get the same result. The center of mass is, therefore, at point O, dividing each of the medians in a ratio of 2: 1, counting from the top. Similar considerations could explain Theorem 4 - the fact that the segments connecting the midpoints of the opposite sides of the quadrilateral divide each other in half (they serve as diagonals of the parallelogram): it is enough to place identical weights at the vertices of the quadrilateral and combine them in pairs in two ways (Fig. 9) ...
Of course, four unit weights located on a plane or in space (at the vertices of a tetrahedron) can be split into two pairs in three ways; the center of mass is located in the middle between the midpoints of the segments connecting these pairs of points (Fig. 10) - an explanation of Theorem 6. (For a flat quadrangle, the result looks like this: two segments connecting the midpoints of opposite sides and the segment connecting the midpoints of the diagonals intersect in one point O and divide it in half).
Through point O - the center of mass of four identical weights - four more segments pass, connecting each of them with the center of mass of the other three. These four segments are divided by point O in a ratio of 3: 1. To explain this fact, you must first find the center of mass of three weights and then attach the fourth.
4. Tetrahedron, octahedron, parallelepiped, cube
At the beginning of the work, we considered a triangle divided by the middle lines into four identical triangles (see Fig. 1). Let's try to do the same construction for an arbitrary triangular pyramid (tetrahedron). We cut the tetrahedron into parts as follows: through the midpoints of three edges emerging from each vertex, we make a flat cut (Fig. 11, a). Then four identical small tetrahedrons will be cut off from the tetrahedron. By analogy with the triangle, one would think that one more tetrahedron of the same type would remain in the middle. But this is not so: the polyhedron, which remains from the large tetrahedron after removing four small ones, will have six vertices and eight faces - it is called an octahedron (Fig. 11.6). It is convenient to test this using a tetrahedral piece of cheese. The resulting octahedron has a center of symmetry, since the midpoints of the opposite edges of the tetrahedron intersect at a common point and are bisected by it.
One interesting construction is connected with the triangle, divided by the middle lines into four triangles: we can consider this figure as a development of some tetrahedron.
Imagine an acute-angled triangle cut out of paper. Having bent it along the middle lines so that the vertices converge at one point, and gluing the edges of the paper converging at this point, we get a tetrahedron, in which all four faces are equal triangles; its opposite edges are equal (fig. 12). Such a tetrahedron is called semiregular. Each of the three "middle sections" of this tetrahedron - parallelograms whose sides are parallel to opposite edges and equal to their halves - will be a rhombus.
Therefore, the diagonals of these parallelograms - three line segments connecting the midpoints of opposite edges - are perpendicular to each other. Among the numerous properties of a semi-regular tetrahedron, we note the following: the sum of the angles converging at each of its vertices is 180 ° (these angles are respectively equal to the angles of the original triangle). In particular, if we start with a sweep in the shape of an equilateral triangle, we get a regular tetrahedron, in which
At the beginning of the work, we saw that each triangle can be viewed as a triangle formed by the midlines of the larger triangle. There is no direct analogy in space for such a construction. But it turns out that any tetrahedron can be regarded as the "core" of a parallelepiped, in which all six edges of the tetrahedron serve as diagonals of the faces. To do this, you need to do the following construction in space. Draw a plane parallel to the opposite edge through each edge of the tetrahedron. The planes drawn through the opposite edges of the tetrahedron will be parallel to each other (they are parallel to the plane of the "middle section" - a parallelogram with vertices at the midpoints of the other four edges of the tetrahedron). Thus, three pairs of parallel planes are obtained, at the intersection of which the desired parallelepiped is formed (two parallel planes intersect the third along parallel lines). The vertices of the tetrahedron serve as four non-adjacent vertices of the constructed parallelepiped (Fig. 13). Conversely, in any parallelepiped, you can select four non-adjacent vertices and cut off corner tetrahedrons from it by planes passing through every three of them. After that, there will be a "core" - a tetrahedron, the edges of which are the diagonals of the parallelepiped faces.
If the original tetrahedron is semiregular, then each face of the constructed parallelepiped will be a parallelogram with equal diagonals, i.e. rectangle.
The converse is also true: the "core" of a rectangular parallelepiped is a semi-regular tetrahedron. Three rhombuses - the middle sections of such a tetrahedron - lie in three mutually perpendicular planes. They serve as planes of symmetry for an octahedron obtained from such a tetrahedron by cutting off the corners.
For a regular tetrahedron, the parallelepiped described around it will be a cube (Fig. 14), and the centers of the faces of this cube - the midpoints of the tetrahedron's edges - will be the vertices of a regular octahedron, all of whose faces are regular triangles. (The three planes of symmetry of the octahedron intersect the tetrahedron in squares.)
Thus, in Figure 14 we see three of the five Platonic solids at once ( regular polyhedra) - cube, tetrahedron and octahedron.
Middle lines of geometric shapes
1. Properties of the middle lines
1. Properties of the triangle:
· When all three middle lines are drawn, 4 equal triangles are formed, similar to the original one with a coefficient of 1/2.
· The middle line is parallel to the base of the triangle and is equal to half of it;
· The middle line cuts off a triangle that is similar to this one, and its area is equal to one quarter of its area.
2. Properties of the quadrangle:
· If in a convex quadrilateral the middle line makes equal angles with the diagonals of the quadrilateral, then the diagonals are equal.
· The length of the midline of the quadrilateral is less than half the sum of the other two sides or equal to it if these sides are parallel, and only in this case.
· The midpoints of the sides of an arbitrary quadrilateral - the vertices of the parallelogram. Its area is half the area of a quadrilateral, and its center lies at the intersection of the midlines. This parallelogram is called the Varignon parallelogram;
· The point of intersection of the midlines of the quadrilateral is their common midpoint and halves the segment connecting the midpoints of the diagonals. In addition, it is the centroid of the vertices of the quadrilateral.
3. Properties of the trapezoid:
· The middle line is parallel to the bases of the trapezoid and is equal to their half-sum;
· The midpoints of the sides of an isosceles trapezoid are the vertices of a rhombus.
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