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 the 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 coefficient of 1/2; its area is equal to one-fourth the area of the original triangle.
- the three middle lines divide the original triangle into four equal triangles. The central of these triangles is called the complementary or medial 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 to a point lying on the other side of the triangle, then this is the midline.
Midline of a quadrilateral
Midline of a quadrilateral- a segment connecting the midpoints of opposite sides of a quadrilateral.
Properties
The first line connects 2 opposite sides. The second connects the other 2 opposite sides. The third connects the centers of two diagonals (not in all quadrilaterals the diagonals are divided in half at the point of intersection).
- 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 midline of a 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 a parallelogram. Its area is equal to half the area of the quadrilateral, and its center lies at the point of intersection of the middle lines. This parallelogram is called the Varignon parallelogram;
- The last point means the following: In a convex quadrilateral you can draw four midlines of the second kind. Midlines of the second kind- four segments inside a quadrilateral, passing through the midpoints of its adjacent sides parallel to the diagonals. Four midlines of the second kind of a convex quadrilateral, cut it into four triangles and one central quadrilateral. This central quadrilateral is a Varignon parallelogram.
- The point of intersection of the midlines of a quadrilateral is their common midpoint and bisects 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 vector of the middle line is equal to half the sum of the vectors of the bases.
Midline of trapezoid
Midline of trapezoid
Midline of trapezoid- a segment connecting the midpoints of the 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 using the formula: E F = A D + B C 2 (\displaystyle EF=(\frac (AD+BC)(2))), Where AD And B.C.- the base of the trapezoid.
Middle line of the triangle
Properties
- The middle line of the triangle is parallel to the third side and equal to half of it.
- when all three middle lines are drawn, 4 equal triangles are formed, similar (even homothetic) to the original one with a coefficient of 1/2.
- the middle line cuts off a triangle that is similar to this one, and its area is equal to one quarter of the area of the original triangle.
Midline of the quadrilateral
Midline of the quadrilateral- a segment connecting the midpoints of opposite sides of a quadrilateral.
Properties
The first line connects 2 opposite sides. The second connects the other 2 opposite sides. The third connects the centers of two diagonals (not all quadrilaterals have centers that intersect)
- 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 midline of a 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 a parallelogram. Its area is equal to half the area of the quadrilateral, and its center lies at the point of intersection of the middle lines. This parallelogram is called the Varignon parallelogram;
- The point of intersection of the midlines of a quadrilateral is their common midpoint and bisects 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 vector of the middle line is equal to half the sum of the vectors of the bases.
Midline of trapezoid
Midline of trapezoid- a segment connecting the midpoints of the sides of this trapezoid. The segment connecting the midpoints of the bases of the trapezoid is called the second midline of the trapezoid.
Properties
- the middle line is parallel to the bases and equal to their half-sum.
See also
Notes
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See what “Midline” is in other dictionaries:
MIDDLE LINE- (1) a trapezoid segment connecting the midpoints of the lateral sides of the trapezoid. The midline of the trapezoid is parallel to its bases and equal to their half-sum; (2) of a triangle, a segment connecting the midpoints of two sides of this triangle: the third side in this case... ... Big Polytechnic Encyclopedia
A triangle (trapezoid) is a segment connecting the midpoints of two sides of a triangle (sides of a trapezoid)... Big Encyclopedic Dictionary
midline- 24 center line: An imaginary line passing through the thread profile so that the thickness of the shoulder is equal to the width of the groove. Source … Dictionary-reference book of terms of normative and technical documentation
Triangle (trapezoid), a segment connecting the midpoints of two sides of the triangle (sides of the trapezoid). * * * MIDDLE LINE MIDDLE LINE of a triangle (trapezoid), a segment connecting the midpoints of two sides of the triangle (lateral sides of the trapezoid) ... Encyclopedic Dictionary
midline- vidurio linija statusas T sritis Kūno kultūra ir sportas apibrėžtis 3 mm linija, dalijanti teniso became paviršių išilgai pusiau. atitikmenys: engl. center line; midtrack line vok. Mittellinie, f rus. middle line...Sporto terminų žodynas
midline- vidurio linija statusas T sritis Kūno kultūra ir sportas apibrėžtis Linija, dalijanti fechtavimosi kovos takelį į dvi lygias dalis. atitikmenys: engl. center line; midtrack line vok. Mittellinie, f rus. middle line…Sporto terminų žodynas
midline- vidurio linija statusas T sritis Kūno kultūra ir sportas apibrėžtis Linija, dalijanti sporto aikšt(el)ę pusiau. atitikmenys: engl. center line; midtrack line vok. Mittellinie, f rus. middle line…Sporto terminų žodynas
1) S. l. triangle, a segment connecting the midpoints of two sides of a triangle (the third side is called the base). S. l. of the triangle is parallel to the base and equal to half of it; area of the parts of the triangle into which c divides it. l.,... ... Great Soviet Encyclopedia
A segment of a triangle connecting the midpoints of two sides of the triangle. The third side of the triangle is called the base of the triangle. S. l. of a triangle is parallel to the base and equal to half its length. In any triangle S. l. cuts off from... ... Mathematical Encyclopedia
Triangle (trapezoid), a segment connecting the midpoints of two sides of the triangle (sides of the trapezoid) ... Natural science. Encyclopedic Dictionary
Gomel Scientific and Practical Conference of Schoolchildren on Mathematics, Its Applications and Information Technologies “Search”
Educational and research work
Center lines of geometric shapes
Morozova Elizaveta
Gomel 2010
Introduction
1.Properties of midlines
2. Triangle, quadrilateral, parallelogram
3. Quadrilateral, tetrahedron. Centers of mass
4. Tetrahedron, octahedron, parallelepiped, cube
Conclusion
List of used literature
Application
Introduction
Geometry is an integral part of general culture, and geometric methods serve as a tool for understanding the world, contribute to the formation of scientific ideas about the surrounding space, and the discovery of the harmony and perfection of the Universe. Geometry starts with a triangle. For two millennia now, the triangle has been a symbol of geometry, but it is not a symbol. A triangle is an atom of geometry. The triangle is inexhaustible - its new properties are constantly being discovered. To talk about all its known properties, you need a volume comparable in volume to that of the Great Encyclopedia. We want to talk about the midlines of geometric shapes and their properties.
Our work traces a chain of theorems that covers the entire geometry course. It starts with the theorem about the midlines of a triangle and leads to interesting properties of the tetrahedron and other polyhedra.
The midline of a figure is a segment connecting the midpoints of two sides of a given figure.
1. Properties of midlines
Properties of a 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 equal to its half;
the middle line cuts off a triangle that is similar to this one, and its area is one quarter of its area.
Properties of a quadrilateral:
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 midline of a 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 a parallelogram. Its area is equal to half the area of the quadrilateral, and its center lies at the point of intersection of the middle lines. This parallelogram is called Varignon's parallelogram;
The point of intersection of the midlines of a quadrilateral is their common midpoint and bisects the segment connecting the midpoints of the diagonals. In addition, it is the centroid of the vertices of the quadrilateral.
Trapezoid properties:
the middle line is parallel to the bases of the trapezoid and equal to their half-sum;
The midpoints of the sides of an isosceles trapezoid are the vertices of a rhombus.
2. Triangle, quadrilateral, parallelogram
To any triangle KLM, three equal triangles AKM, BLK, CLM can be attached, each of which, together with the triangle KLM, forms a parallelogram (Fig. 1). In this case, AK = ML = KB, and the vertex K is adjacent to three angles equal to three different angles of the triangle, totaling 180°, therefore K is the middle of the segment AB; similarly, L is the midpoint of the segment BC, and M is the midpoint of the segment CA.
Theorem 1. If we connect the midpoints of the sides in any triangle, we get four equal triangles, with the middle one forming a parallelogram with each of the other three.
This formulation involves all three middle lines of the triangle 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 equal to half of it (see Fig. 1).
It is this theorem and its converse - that a straight line parallel to the base and passing through the middle of one side of a triangle divides the other side in half - are most often needed when solving problems.
From the theorem on the midlines of a triangle follows the property of the midline of a trapezoid (Fig. 2), as well as the theorems on segments connecting the midpoints of the sides of an arbitrary quadrilateral.
Theorem 3. The midpoints of the sides of a quadrilateral are the vertices of a 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.
In fact, if K and L are the midpoints of sides AB and BC (Fig. 3), then KL is the midline of triangle ABC, therefore the segment KL is parallel to the diagonal AC and equal to half of it; if M and N are the midpoints of the 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 an interesting fact (Part 4).
Theorem 4. In any quadrilateral, the segments connecting the midpoints of opposite sides are divided in half 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 in half by the point of intersection (this point is the center of symmetry of the parallelogram).
We see that Theorems 3 and 4 and our reasoning remain true both for a non-convex quadrilateral and for a self-intersecting quadrilateral closed broken line (Fig. 4; in the latter case it may turn out that the parallelogram KLMN is “degenerate” - points K, L, M, N lie on the same straight line).
Let us show how from Theorems 3 and 4 we can derive the main theorem on the medians of a triangle.
Theorem5 . The medians of a 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 their intersection point. The midpoints of the sides of a non-convex quadrilateral ABCO are the points K, L, M and N (Fig. 5) - 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 ratio of 2:1.
Instead of the median SC, we could consider the median drawn from vertex B and make sure in the same way that it divides the median AL in the ratio 2:1, that is, 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 broken line consisting of four links AB, BC, CD, DA, whose four vertices A, B, C, D do not lie in the same plane.
Such a spatial quadrilateral 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 midlines KL and MN of triangles ABC and ADC remain their midlines and will be parallel to the segment AC and equal to AC/2. (Here we use the fact that the basic 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 vertices of the parallelogram; Thus, the segments KM and LN intersect and are divided in half by the intersection point. Instead of a quadrilateral, 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. By cutting the tetrahedron along this plane (Fig. 6, b), we obtain a parallelogram KLMN, two sides of which are parallel to the edge AC and equal
AC/2, and the other two are parallel to 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 corollary:
Theorem 6. Three segments connecting the midpoints of opposite edges of the tetrahedron intersect at one point and are divided in half by it (Fig. 7).
This and other facts discussed above are naturally explained in the language of mechanics - using the concept of the center of mass. Theorem 5 talks about 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 a tetrahedron. These points are the centers of mass of the triangle and tetrahedron, respectively. Let us first return to Theorem 5 on medians.
Let's place three identical weights at the vertices of the triangle (Fig. 8).
Let's take the mass of each as one. Let's find the center of mass of this load system.
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 these weights can be replaced by 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 loads: 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 SC in the ratio 2:1 (closer to the load at point K with a larger 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 segment BC with the load at point A. Or first combine the loads A and C, a. then add B. Either way we should get the same result. The center of mass is thus located at point O, dividing each of the medians in a ratio of 2:1, counting from the vertex. Similar considerations could explain Theorem 4 - the fact that the segments connecting the midpoints of opposite sides of a quadrilateral bisect each other (serve as diagonals of a 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 divided 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) - explanation of Theorem 6. (For a flat quadrilateral, the result obtained 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 divide it in half).
Through point O - the center of mass of four identical loads - 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 the three weights and then attach the fourth.
4. Tetrahedron, octahedron, parallelepiped, cube
At the beginning of the work, we looked at 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). Let's cut the tetrahedron into pieces as follows: through the middles of the three edges emerging from each vertex, we draw a flat cut (Fig. 11, a). Then four identical small tetrahedrons will be cut from the tetrahedron. By analogy with a triangle, one would think that there would be another similar tetrahedron in the middle. But this is not so: the polyhedron that remains from the large tetrahedron after removing the four small ones will have six vertices and eight faces - it is called an octahedron (Fig. 11.6). A convenient way to test this is by using a piece of cheese in the shape of a tetrahedron. 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 associated with a triangle divided by the middle lines into four triangles: we can consider this figure as the development of a certain tetrahedron.
Let's imagine an acute triangle cut out of paper. By bending 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 semi-regular. 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 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 equal to 180° (these angles are respectively equal to the angles of the original triangle). In particular, if we start with an equilateral triangle, we get a regular tetrahedron with
At the beginning of the work we saw that each triangle can be considered as a triangle formed by the midlines of a 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. Through each edge of the tetrahedron we draw a plane parallel to the opposite edge. 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 in the middle of the other four edges of the tetrahedron). This produces three pairs of parallel planes, the intersection of which forms the desired parallelepiped (two parallel planes are intersected by a third along parallel straight lines). The vertices of the tetrahedron serve as four non-adjacent vertices of the constructed parallelepiped (Fig. 13). On the contrary, in any parallelepiped you can select four non-adjacent vertices and cut off corner tetrahedrons from it with planes passing through each three of them. After this, a “core” will remain - a tetrahedron, the edges of which are the diagonals of the faces of the parallelepiped.
If the original tetrahedron is semi-regular, then each face of the constructed parallelepiped will be a parallelogram with equal diagonals, i.e. rectangle.
The opposite 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 of the 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 middles of the edges of the tetrahedron - 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 immediately see three of the five Platonic solids (regular polyhedra) - cube, tetrahedron and octahedron.
Conclusion
Based on the work done, the following conclusions can be drawn:
Midlines have various useful properties in geometric shapes.
One theorem can be proven using the center line of the figures, and also explained in the language of mechanics - using the concept of the center of mass.
Using midlines, you can construct various planimetric (parallelogram, rhombus, square) and stereometric figures (cube, octahedron, tetrahedron, etc.).
The properties of midlines help to rationally solve problems of any level.
List of sources and literature used
Monthly popular science physics and mathematics journal of the USSR Academy of Sciences 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 lessons in geometry, 10th grade: a manual for teachers. - Mn.: TetraSystems, 2004 p. 68.76, 78.
Application
Why can't the middle line of a trapezoid pass through the intersection point 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 ribs AD and DC, respectively. Is it true that lines EF and CT are parallel?
In a triangular prism ABCA 1 B 1 C 1 points O and F are the middle of the edges AB and BC, respectively. Points T and K are the middle of segments AB 1 and BC 1, respectively. How are the direct lines TK and OF 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 middle of edge CC 1, and point F lies on 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 is formed by a closed line. There are convex and non-convex. U figures there are sides..., sector, sphere, segment, sine, middle, average line, ratio, property, degree, stereometry, secant...
Definition
A parallelogram is a quadrilateral whose opposite sides are parallel in pairs.
Theorem (first sign of a parallelogram)
If two sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram.
Proof
Let the sides \(AB\) and \(CD\) be parallel in the quadrilateral \(ABCD\) and \(AB = CD\) .
Let's draw a diagonal \(AC\) dividing this quadrilateral into two equal triangles: \(ABC\) and \(CDA\) . These triangles are equal in two sides and the angle between them (\(AC\) is the common side, \(AB = CD\) by condition, \(\angle 1 = \angle 2\) as crosswise angles at the intersection of parallel lines \ (AB\) and \(CD\) secant \(AC\) ), so \(\angle 3 = \angle 4\) . But the angles \(3\) and \(4\) lie crosswise at the intersection of the lines \(AD\) and \(BC\) by the secant \(AC\), therefore, \(AD\parallel BC\) . Thus, in the quadrilateral \(ABCD\) the opposite sides are pairwise parallel, and, therefore, the quadrilateral \(ABCD\) is a parallelogram.
Theorem (second sign of a parallelogram)
If in a quadrilateral the opposite sides are equal in pairs, then this quadrilateral is a parallelogram.
Proof
Let's draw a diagonal \(AC\) of this quadrilateral \(ABCD\) dividing it into triangles \(ABC\) and \(CDA\) .
These triangles are equal on three sides (\(AC\) – common, \(AB = CD\) and \(BC = DA\) by condition), therefore \(\angle 1 = \angle 2\) – lying crosswise at \(AB\) and \(CD\) and secant \(AC\) . It follows that \(AB\parallel CD\) . Since \(AB = CD\) and \(AB\parallel CD\) , then according to the first criterion of a parallelogram, the quadrilateral \(ABCD\) is a parallelogram.
Theorem (third sign of a parallelogram)
If the diagonals of a quadrilateral intersect and are divided in half by the point of intersection, then this quadrilateral is a parallelogram.
Proof
Consider a quadrilateral \(ABCD\) in which the diagonals \(AC\) and \(BD\) intersect at the point \(O\) and are bisected by this point.
Triangles \(AOB\) and \(COD\) are equal according to the first sign of equality of triangles (\(AO = OC\), \(BO = OD\) by condition, \(\angle AOB = \angle COD\) as vertical angles), so \(AB = CD\) and \(\angle 1 = \angle 2\) . From the equality of the angles \(1\) and \(2\) (crosswise lying at \(AB\) and \(CD\) and the secant \(AC\) ) it follows that \(AB\parallel CD\) .
So, in the quadrilateral \(ABCD\) the sides \(AB\) and \(CD\) are equal and parallel, which means that according to the first criterion of a parallelogram, the quadrilateral \(ABCD\) is a parallelogram.
Properties of a parallelogram:
1. In a parallelogram, opposite sides are equal and opposite angles are equal.
2. The diagonals of a parallelogram are divided in half by the point of intersection.
Properties of the bisector of a parallelogram:
1. The bisector of a parallelogram cuts off an isosceles triangle from it.
2. Bisectors of adjacent angles of a parallelogram intersect at right angles.
3. Bisector segments of opposite angles are equal and parallel.
Proof
1) Let \(ABCD\) be a parallelogram, \(AE\) be the bisector of the angle \(BAD\) .
Angles \(1\) and \(2\) are equal, lying crosswise with parallel lines \(AD\) and \(BC\) and the secant \(AE\). Angles \(1\) and \(3\) are equal, since \(AE\) is a bisector. In the end \(\angle 3 = \angle 1 = \angle 2\), which means that the triangle \(ABE\) is isosceles.
2) Let \(ABCD\) be a parallelogram, \(AN\) and \(BM\) be the bisectors of angles \(BAD\) and \(ABC\), respectively.
Since the sum of one-sided angles for parallel lines and a transversal is equal to \(180^(\circ)\), then \(\angle DAB + \angle ABC = 180^(\circ)\).
Since \(AN\) and \(BM\) are bisectors, then \(\angle BAN + \angle ABM = 0.5(\angle DAB + \angle ABC) = 0.5\cdot 180^\circ = 90^(\circ)\), where \(\angle AOB = 180^\circ - (\angle BAN + \angle ABM) = 90^\circ\).
3. Let \(AN\) and \(CM\) be the bisectors of the angles of the parallelogram \(ABCD\) .
Since opposite angles in a parallelogram are equal, then \(\angle 2 = 0.5\cdot\angle BAD = 0.5\cdot\angle BCD = \angle 1\). In addition, the angles \(1\) and \(3\) are equal, lying crosswise with parallel lines \(AD\) and \(BC\) and the secant \(CM\), then \(\angle 2 = \angle 3\) , which implies that \(AN\parallel CM\) . Moreover, \(AM\parallel CN\) , then \(ANCM\) is a parallelogram, hence \(AN = CM\) .
A quadrilateral in which only two sides are parallel is called trapezoid.
The parallel sides of a trapezoid are called its reasons, and those sides that are not parallel are called 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 Trapezoid
The midline is a segment connecting the midpoints of the sides of the trapezoid. The midline of the trapezoid is parallel to its bases.
Theorem:
If the 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 middle line is equal to the arithmetic mean of the lengths of its bases
MN || AB || DCAM = MD; BN=NC
MN midline, AB and CD - bases, AD and BC - lateral sides
MN = (AB + DC)/2
Theorem:
The length of the midline of a trapezoid is equal to the arithmetic mean of the lengths of its bases.
Main task: Prove that the midline of a trapezoid bisects a segment whose ends lie in the middle of the bases of the trapezoid.
Middle Line of the Triangle
The segment connecting the midpoints of two sides of a triangle is called the midline of the triangle. It is parallel to the third side and its length is equal to 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 the triangle, then it bisects the third side.
AM = MC and BN = NC =>
Applying the midline properties of a triangle and trapezoid
Dividing a segment into a certain number of equal parts.
Task: Divide segment AB into 5 equal parts.
Solution:
Let p be a random ray whose origin is point A and which does not lie on line AB. We sequentially set aside 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 that are parallel to A 5 B. They intersect AB respectively at points B 4, B 3, B 2 and B 1. These points divide 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 obtain B 4 B 3 = B 3 B 2
While from the 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 get:
AB 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 = B 4 B
It is clear that to divide the segment AB into another number of equal parts, we need to project the same number of equal segments onto the ray p. And then continue in the manner described above.
