Median lines of the quadrilateral. Parallelogram
107. We know (item 102) that the locus of points equidistant from two given parallel lines is the middle parallel. If in this way AB and CD (Fig. 114) are two parallel ones and MN for them the mean is parallel, then the distances of any point E of this mean parallel from AB and CD are equal to each other, that is, by constructing EF ⊥ AB and EG ⊥ CD , we get that EF = EG.
It is clear that the constructed perpendiculars EF and EG are continuation of each other and form one segment FG, perpendicular to our parallel AB and CD, and this segment is divided by the middle parallel (at point E) in half. So, any segment perpendicular to two parallel and enclosed between them is divided by the middle parallel in half.
The question now arises: will not some segment KL, not perpendicular to AB and CD, also be divided in half by the mean parallel. Let KL intersect MN at the point O. Construct through the point O a segment HI perpendicular to lines AB and CD. Then OH = OI. Since, in addition, ∠HOK = ∠IOL, as vertical, then right-angled triangles OHK and OIL are equal, which implies that OK = OL. So, it turns out that any segment between two parallel ones is divided by the middle parallel in half.
Let AB || CD (black 115). Having constructed between them a series of some segments EF, GH, KI, etc., we, according to the previous one, find that the midpoints of these segments lie on the middle parallel MN. In general, we come to the following conclusion:
The locus of the midpoints of all kinds of segments enclosed between two parallel ones is the middle parallel.
This gives rise to the possibility of various constructions of the mean parallel for two given parallel lines: 1) we can, construct any segment EF, enclosed between two data parallel AB and CD, divide it in half and through its middle build the line MN || AB || CD is a straight line MN and should serve as a middle parallel, and it should halve all possible segments (eg, GH, KI, etc.) between AB and CD. 2) We can build two segments, for example, EH and KI, enclosed between AB and CD, divide each of them in half and build a straight line MN through their midpoints - it should serve as a middle parallel.
108. Let us apply the properties of the mean parallel to the figures we are familiar with, and above all to the triangle.
Suppose we have ∆ABC (drawing 116). Here we do not directly have two parallel ones, but we can always get them, for example, by constructing a straight line EF || through the vertex A. BC (this straight line EF could not have been drawn in the drawing, since it does not play a significant role in the future and since it is enough just to know that it exists). Then we have two parallel BC and EF and two line segments AB and AC, enclosed between them. Dividing them in half at the points M and N (AM = MB and AN = NC) and constructing the line MN through M and N, we obtain the mean parallel MN, that is, MN || BC (and || EF, but this is not essential for us). From this we conclude:
the line connecting the midpoints of the two sides of the triangle is parallel to its third side.
The segment connecting the midpoints of the two sides of the triangle is called the middle line of the triangle... So, we have a segment MN is the middle line of our triangle.
Suppose we have ∆ABC (drawing 117). Divide each of its sides in half: let M be the midpoint of AB (w. AM = MB), N the midpoint of AC (AN = NC), and P the midpoint of BC (BP = PC); connect points M, N and P with segments MN, MP and PN, - each of these segments is the middle line for our triangle. Thus, there are three middle lines in the triangle.
According to the previous one, we will have: MN || BC, MP || AC and NP || AB. Therefore AMPN, BMNP and PMNC are parallelograms. Since in the parallelogram the opposite sides are equal, we have: MN = BP (from the BMNP parallelogram), but BP = BC / 2 (because point P is the middle of BC); therefore MN = BC / 2. Also from the parallelogram AMPN we get: MP = AN = AC / 2 and from the parallelogram AMPN - PN = AM = AB / 2. Hence we conclude:
each middle line of a triangle connecting the midpoints of its two sides is parallel to the third and equal to its half.
109. Let us now turn to quadrangles and first dwell on those quadrangles, whose two sides are parallel. It is customary to call such quadrangles trapezoids. On black. 118 shows two different kinds trapezoid: 1) trapezoid ABCD, where BC || AD, but AB is not parallel to CD, - this trapezoid has an area (see item 79) and 2) trapezoid A "B" C "D", where A "D" || B "C" - this trapezoid has no area (p. 79).
Let us first consider the trapezoid ABCD (black. 118 bis), which has an area. Here BD || AD. Therefore, we have two parallel BC and AD and between them the segments AB and CD. Dividing these segments in half at points M and N (AM = MB and CN = ND) and connecting them with a straight line MN, we obtain the average parallel MN for BC and AD, that is, MN || BC || AD. The segment MN of this straight line is called the middle line of the trapezoid (it should be added: “connecting the midpoints of the non-parallel sides”, because in the trapezoid, as in any quadrilateral, 6 midlines can be considered, which is the case in paragraph 110). So we got that MN || BC || AD. Further, having constructed the diagonal AC, we obtain another third segment AC, enclosed between parallel BC and AD - its midpoint must lie (item 107) on the middle parallel, i.e., the point P, where MN and AC intersect, is the midpoint of the segment AC. Therefore, MP is the midline of triangle ABC and PN is the midline of ∆ACD. Based on the previous one, we have: MP = BC / 2 and PN = AD / 2. From here we get: MN = MP + PN = BC / 2 + AD / 2 or MN = (BC + AD) / 2. So,
the middle line connecting the midpoints of the non-parallel sides of a trapezoid having an area is parallel to its parallel sides and is equal to their half-sum.
Now suppose we have a trapezoid ABCD (Fig. 118 bis), which has no area. Here also BC || AD and therefore the midpoints M and N of sides AB and CD lie on the middle parallel, that is, here we also have: MN || BC || AD. Having constructed the diagonal AC, we get a segment AC, enclosed between parallel BC and AD, and its midpoint, point P, must lie on the middle parallel. Therefore PM is the midline of triangle ABC and hence PM = BC / 2; also PN is the middle line ∆ABC and, next, PN = AD / 2. Since MN = PN - PM, we get MN = PN - PM = AD / 2 - BC / 2 or MN = (AD - BC) / 2. So,
the middle line connecting the midpoints of the non-parallel sides of a trapezoid that has no area is parallel to its parallel sides and is equal to their half-difference.
110. Suppose we have some quadrangle ABCD (having area) - (Fig. 119). Find the midpoints M, N, P and Q of its sides and connect them in pairs. We get 6 middle lines of the quadrilateral.
Here are the properties of these middle lines.
1) The midlines connecting the midpoints of the successive sides of the quadrilateral form a parallelogram.
To clarify this property, we construct a diagonal AC. Then from ∆ABC we have (item 108) MN || AC and from ∆ACD on the same basis: PQ || AC, - next, MN || PQ. Having constructed another diagonal BD, we find with its help that NP || MQ, therefore, MNPQ is a parallelogram.
2) The midlines of the quadrangle, connecting the midpoints of opposite sides, are mutually halved.
This property is now obvious, since MP and NQ are parallelogram diagonals.
The intersection point O of lines MP and NQ also includes lines connecting the midpoints of the diagonals AC and BD (the diagonal BD is not shown in the drawing). This follows from the fact that AC AND BD are sides of an arealess quadrilateral ACBD, to which everything stated at the beginning of this clause applies.
111. We were able (pp. 57, 59) to divide the segment in half and, consequently, by 4, by 8, and generally into 2n equal parts. Now we can divide this segment into 3, 5 and in general into as many equal parts as we like.
Let, for example, it is required to divide the segment AB (black. 120) into 5 equal parts. Let us construct an arbitrary line AC through point A (forming an angle with AB that is different from the rectified one) and put on AC five arbitrary, but equal, segments AE = EF = FG = GH = HO. Construct line OB and, through points E, F, G and H, construct lines EE ", FF", GG ", HH", parallel to OB.
Consider ∆AFF ", since AE = EF, then E is the midpoint of side AF and EE" (she || FF ") is the midline of this triangle, therefore, AE" = E "F".
Consider then the trapezoid EE "G" G. Since EF = FG, FF "|| EE", then FF "is the middle line of the trapezoid EE" GG ", hence E" F "= F" G ". We also find that GG" is the middle line of the trapezoid FF " H "H and, next, F" G "= G" H ", etc. Combining the obtained equalities, we find AE" = E "F" = F "G" = G "H" = H "B", that is, the segment AB is divided into 5 equal parts.
The conclusion can be drawn from the solution of this problem:
If on one side of the corner we lay aside equal segments and build a series of parallel straight lines through their ends, then on the other side of the corner we will get equal segments.
Adding... We laid off equal segments on one straight line in a row, starting from the point of intersection of two straight lines (AB and AC of drawing 120), but it is possible to arrive at the same result with a different method of depositing equal segments. Drawing 120 bis gives two variants of such a construction: on line AD (see drawing 120 bis on the left or on the right) we set aside two equal segments AB and CD and construct parallel AA "|| BB" || CC "|| DD". Then take point O, the midpoint of segment BC, and construct OO "|| BB" || CC "|| AA" || DD ". Then OO" is the middle line of the trapezoid BCC "B"; therefore B "O" = O "C (p. 109). Since AB = CD and BO = OC, then AO is also = OD; therefore OO" is also the middle line of trapezoid ADD "A" (in the drawing on the right, this trapezoid is ADD "A" - not having area, see item 109) - and also A "O" = O "D". Hence we have A "O" - B "O" = O "D" - O "C" (because both the deducted and deducted of both differences are equal), or A "B" = C "D". Other combinations are also possible (for example, move the neg. CD of the right figure so that point C is to the right of the point of intersection of lines AD and A "D"). The general conclusion is as follows: if two straight lines are constructed, on one of them two equal segments are laid off in some way and parallel ones are constructed through their ends, then these latter will select two equal segments on the other line.
112. Exercises.
- Through the vertices of this triangle, straight lines are drawn that are parallel to its sides. Show that the new triangle has sides twice as many as the sides of the given one, and that the vertices of the given one are the midpoints of the sides of the new one (cf. exercise 7 from p. 54).
- Construct a triangle if given the midpoints of its three sides.
- Construct a parallelogram if given the midpoints of its three sides.
- It is known (item 110) that the midpoints of the four sides of a quadrilateral are the vertices of a parallelogram. When does this parallelogram turn into a rhombus, when into a rectangle, when into a square?
- The straight line connecting the apex of the triangle with the middle of the opposite side (median) and the straight line connecting the midpoints of the other two sides of the triangle are mutually divided in half.
- Extend one side of the triangle to a segment equal to this side, and connect the end of the segment to the middle of the other side. The last connecting line cuts off a segment equal to 1/3 of this side from the third side of the triangle. (Construct another straight line parallel to the last connecting straight line through the apex of the triangle, opposite to the side that was continued).
- If on side AB of parallelogram ABCD we put off the segment AM = (1 / n) AB (e.g. (1/7) AB) and connect D with M, then DM intersects the diagonal AC at point N so that AN = (1 / ( n + 1)) AC (in the example taken (1/8) AC).
To find out, it is necessary to postpone BM "= AM on the extension of side AB and connect C to M"; then C "M" || DM, - take paragraph 111.
Middle lines of quadrangles and their properties Completed by: Dmitry Matveev Teacher: Rychkova Tatyana Viktorovna Lyceum "Dubna" 9IM 2007 Average lines and Parallelogram Varignon Other properties midline 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 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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Educational research work
Middle lines of 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 of the triangle, are adjacent to the vertex 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 quadrangle.
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 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 SC, we could consider the median drawn from the vertex B, and make sure in the same way that it also 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 their middle lines and will be parallel to the AC segment 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 segment AB, so that these weights can be replaced with one weight of mass 2, placed in the middle K of 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 located 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 considered 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 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 ...
A quadrilateral with only two sides parallel is called trapezoid.
The parallel sides of the trapezoid are called it grounds, and those sides that are not parallel are called lateral sides... If the sides are equal, then such a trapezoid is isosceles. The distance between the bases is called the height of the trapezoid.
Middle Line of Trapezium
The midline is the line segment connecting the midpoints of the sides of the trapezoid. The middle line of the trapezoid is parallel to its bases.
Theorem:
If a straight line crossing the middle of one side is parallel to the bases of the trapezoid, then it bisects the second side of the trapezoid.
Theorem:
The length of the midline is equal to the arithmetic mean of the lengths of its bases
MN || AB || DCAM = MD; BN = NC
MN middle line, AB and CD - bases, AD and BC - sides
MN = (AB + DC) / 2
Theorem:
The length of the midline of the trapezoid is equal to the arithmetic mean of the lengths of its bases.
The main task: Prove that the middle line of a trapezoid bisects a segment whose ends lie in the middle of the base of the trapezoid.
Center Line of the Triangle
The segment connecting the midpoints of the two sides of the triangle is called the midline of the triangle. It is parallel to the third side and is half the length of the third side.
Theorem: If a line intersecting the midpoint of one side of a triangle is parallel to the other side of this triangle, then it divides the third side in half.
AM = MC and BN = NC =>
Applying Triangle and Trapezoid Midline Properties
Division of a segment into a certain number of equal parts.
Task: Divide segment AB into 5 equal parts.
Solution:
Let p be a random ray with origin at point A and not lying on line AB. We successively lay 5 equal segments on p AA 1 = A 1 A 2 = A 2 A 3 = A 3 A 4 = A 4 A 5
We connect A 5 to B and draw such lines through A 4, A 3, A 2 and A 1, which are parallel to A 5 B. They intersect AB, respectively, at points B 4, B 3, B 2 and B 1. These points divide line segment AB into 5 equal parts. Indeed, from the trapezoid BB 3 A 3 A 5 we see that BB 4 = B 4 B 3. In the same way, from the trapezoid B 4 B 2 A 2 A 4 we get B 4 B 3 = B 3 B 2
While from a trapezoid B 3 B 1 A 1 A 3, B 3 B 2 = B 2 B 1.
Then from B 2 AA 2 it follows that B 2 B 1 = B 1 A. In conclusion, we obtain:
AB 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 = B 4 B
It is clear that in order to divide segment AB into another number of equal parts, we need to project the same number of equal segments onto ray p. And then continue in the way described above.
