What is the length of the midline? The middle line of the triangle. How to find the midline of a triangle
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How to find the midpoint of a triangle: a geometry problem. The main elementary problems in Euclidean geometry came to us from antiquity. They contain the primary essence itself and the necessary basic knowledge about human perception of spatial forms. One such problem is the problem of finding the midpoint of a triangle. Today, this problem is considered as an educational technique for developing the intellectual abilities of schoolchildren. In the ancient world, knowledge of how to find the middle of a triangle was also used in practice: in land management, in the manufacture of various mechanisms, etc. What is the essence of this geometric rebus? What is the median? Before solving the problem, you need to familiarize yourself with the simplest geometric terminology regarding triangles. First of all, each triangle has three vertices, three sides and three angles, which is where the name of this triangle comes from. geometric figure
. It is important to know what the lines connecting vertices to opposite sides are called: height, bisector and median.
Height is a line perpendicular to the side opposite the vertex from which it is drawn; bisector - divides an angle in half; The median divides the side opposite to the outgoing vertex in half. To solve this problem, you need to know how to find the coordinates of the midpoint of a segment, because it is the point of intersection of the medians of the triangle that is its midpoint.
Find the midpoints of the sides of the triangle. Finding the midpoint of a segment is also a classic geometric problem, to solve which you will need a compass and a ruler without divisions. We place the needle of the compass at the end point of the segment and draw a semicircle larger than half of the segment in the middle of the last one. We do the same on the other side of the segment. The resulting semicircles will necessarily intersect at two points, because their radii are greater than half of the original segment.
We build the middle of the triangle. By connecting the vertices of the triangle with the midpoints of the opposite sides with straight lines, we obtain three medians. This may surprise some, but one of the laws of harmony of this geometric figure is that all three medians always intersect at one point. It is this point that will be the desired midpoint of the triangle, which is not so difficult to find if you know how to construct the midpoint of the segment.
It is also interesting that the point of intersection of the medians represents not only the geometric, but also the “physical” middle of the triangle. That is, if, for example, you cut a triangle out of plywood, find its middle and place this point on the tip of the needle, then ideally such a figure will balance and not fall. Elementary geometry contains many such fascinating “secrets”, the knowledge of which helps to comprehend the harmony of the surrounding world and the nature of more complex things.
\[(\Large(\text(Similarity of triangles)))\]
Definitions
Two triangles are called similar if their angles are respectively equal and the sides of one triangle are proportional to the similar sides of the other
(sides are called similar if they lie opposite equal angles).
The coefficient of similarity of (similar) triangles is a number equal to the ratio of the similar sides of these triangles.
Definition
The perimeter of a triangle is the sum of the lengths of all its sides.
Theorem
The ratio of the perimeters of two similar triangles is equal to the similarity coefficient.
Proof
Consider triangles \(ABC\) and \(A_1B_1C_1\) with sides \(a,b,c\) and \(a_1, b_1, c_1\) respectively (see figure above).
Then \(P_(ABC)=a+b+c=ka_1+kb_1+kc_1=k(a_1+b_1+c_1)=k\cdot P_(A_1B_1C_1)\)
Theorem
The ratio of the areas of two similar triangles is equal to the square of the similarity coefficient.
Proof
Let the triangles \(ABC\) and \(A_1B_1C_1\) be similar, and \(\dfrac(AB)(A_1B_1) = \dfrac(AC)(A_1C_1) = \dfrac(BC)(B_1C_1) = k\). Let us denote by the letters \(S\) and \(S_1\) the areas of these triangles, respectively.
Since \(\angle A = \angle A_1\) , then \(\dfrac(S)(S_1) = \dfrac(AB\cdot AC)(A_1B_1\cdot A_1C_1)\)(by the theorem on the ratio of the areas of triangles having equal angles).
Because \(\dfrac(AB)(A_1B_1) = \dfrac(AC)(A_1C_1) = k\), That \(\dfrac(S)(S_1) = \dfrac(AB)(A_1B_1)\cdot\dfrac(AC)(A_1C_1) = k\cdot k = k^2\), which was what needed to be proven.
\[(\Large(\text(Signs of similarity of triangles)))\]
Theorem (the first sign of similarity of triangles)
If two angles of one triangle are respectively equal to two angles of another triangle, then such triangles are similar.
Proof
Let \(ABC\) and \(A_1B_1C_1\) be triangles such that \(\angle A = \angle A_1\) , \(\angle B = \angle B_1\) . Then, by the theorem on the sum of angles of a triangle \(\angle C = 180^\circ - \angle A - \angle B = 180^\circ - \angle A_1 - \angle B_1 = \angle C_1\), that is, the angles of the triangle \(ABC\) are respectively equal to the angles of the triangle \(A_1B_1C_1\) .
Since \(\angle A = \angle A_1\) and \(\angle B = \angle B_1\) , then \(\dfrac(S_(ABC))(S_(A_1B_1C_1)) = \dfrac(AB\cdot AC)(A_1B_1\cdot A_1C_1)\) And \(\dfrac(S_(ABC))(S_(A_1B_1C_1)) = \dfrac(AB\cdot BC)(A_1B_1\cdot B_1C_1)\).
From these equalities it follows that \(\dfrac(AC)(A_1C_1) = \dfrac(BC)(B_1C_1)\).
Similarly, it is proved that \(\dfrac(AC)(A_1C_1) = \dfrac(AB)(A_1B_1)\)(using equalities \(\angle B = \angle B_1\) , \(\angle C = \angle C_1\) ).
As a result, the sides of the triangle \(ABC\) are proportional to the similar sides of the triangle \(A_1B_1C_1\), which is what needed to be proven.
Theorem (second criterion for the similarity of triangles)
If two sides of one triangle are proportional to two sides of another triangle and the angles between these sides are equal, then the triangles are similar.
Proof
Consider two triangles \(ABC\) and \(A"B"C"\) such that \(\dfrac(AB)(A"B")=\dfrac(AC)(A"C")\), \(\angle BAC = \angle A"\) Let us prove that the triangles \(ABC\) and \(A"B"C"\) are similar. Taking into account the first sign of similarity of triangles, it is enough to show that \(\angle B = \angle B"\) .
Consider a triangle \(ABC""\) with \(\angle 1 = \angle A"\) , \(\angle 2 = \angle B"\) . Triangles \(ABC""\) and \(A"B"C"\) are similar according to the first criterion of similarity of triangles, then \(\dfrac(AB)(A"B") = \dfrac(AC"")(A"C")\).
On the other hand, by condition \(\dfrac(AB)(A"B") = \dfrac(AC)(A"C")\). From the last two equalities it follows that \(AC = AC""\) .
Triangles \(ABC\) and \(ABC""\) are equal in two sides and the angle between them, therefore, \(\angle B = \angle 2 = \angle B"\).
Theorem (third sign of similarity of triangles)
If three sides of one triangle are proportional to three sides of another triangle, then the triangles are similar.
Proof
Let the sides of the triangles \(ABC\) and \(A"B"C"\) be proportional: \(\dfrac(AB)(A"B") = \dfrac(AC)(A"C") = \dfrac(BC)(B"C")\). Let us prove that the triangles \(ABC\) and \(A"B"C"\) are similar.
To do this, taking into account the second criterion for the similarity of triangles, it is enough to prove that \(\angle BAC = \angle A"\) .
Consider a triangle \(ABC""\) with \(\angle 1 = \angle A"\) , \(\angle 2 = \angle B"\) .
Triangles \(ABC""\) and \(A"B"C"\) are similar according to the first criterion of similarity of triangles, therefore, \(\dfrac(AB)(A"B") = \dfrac(BC"")(B"C") = \dfrac(C""A)(C"A")\).
From the last chain of equalities and conditions \(\dfrac(AB)(A"B") = \dfrac(AC)(A"C") = \dfrac(BC)(B"C")\) it follows that \(BC = BC""\) , \(CA = C""A\) .
Triangles \(ABC\) and \(ABC""\) are equal on three sides, therefore, \(\angle BAC = \angle 1 = \angle A"\).
\[(\Large(\text(Thales' Theorem)))\]
Theorem
If you mark equal segments on one side of an angle and draw parallel straight lines through their ends, then these straight lines will also cut off equal segments on the other side.
Proof
Let's prove first lemma: If in \(\triangle OBB_1\) a straight line \(a\parallel BB_1\) is drawn through the middle \(A\) of side \(OB\), then it will also intersect side \(OB_1\) in the middle.
Through the point \(B_1\) we draw \(l\parallel OB\) . Let \(l\cap a=K\) . Then \(ABB_1K\) is a parallelogram, therefore \(B_1K=AB=OA\) and \(\angle A_1KB_1=\angle ABB_1=\angle OAA_1\); \(\angle AA_1O=\angle KA_1B_1\) like vertical. So, according to the second sign \(\triangle OAA_1=\triangle B_1KA_1 \Rightarrow OA_1=A_1B_1\). The lemma is proven.
Let's move on to the proof of the theorem. Let \(OA=AB=BC\) , \(a\parallel b\parallel c\) and we need to prove that \(OA_1=A_1B_1=B_1C_1\) .
Thus, according to this lemma \(OA_1=A_1B_1\) . Let's prove that \(A_1B_1=B_1C_1\) . Let us draw a straight line \(d\parallel OC\) through the point \(B_1\), and let \(d\cap a=D_1, d\cap c=D_2\) . Then \(ABB_1D_1, BCD_2B_1\) are parallelograms, therefore, \(D_1B_1=AB=BC=B_1D_2\) . Thus, \(\angle A_1B_1D_1=\angle C_1B_1D_2\) like vertical \(\angle A_1D_1B_1=\angle C_1D_2B_1\) lying like crosses, and, therefore, according to the second sign \(\triangle A_1B_1D_1=\triangle C_1B_1D_2 \Rightarrow A_1B_1=B_1C_1\).
Thales' theorem
Parallel lines cut off proportional segments on the sides of an angle.
Proof
Let parallel lines \(p\parallel q\parallel r\parallel s\) divided one of the lines into segments \(a, b, c, d\) . Then the second straight line should be divided into segments \(ka, kb, kc, kd\), respectively, where \(k\) is a certain number, the same proportionality coefficient of the segments.
Let us draw through the point \(A_1\) a line \(p\parallel OD\) (\(ABB_2A_1\) is a parallelogram, therefore, \(AB=A_1B_2\) ). Then \(\triangle OAA_1 \sim \triangle A_1B_1B_2\) at two corners. Hence, \(\dfrac(OA)(A_1B_2)=\dfrac(OA_1)(A_1B_1) \Rightarrow A_1B_1=kb\).
Similarly, we draw a straight line through \(B_1\) \(q\parallel OD \Rightarrow \triangle OBB_1\sim \triangle B_1C_1C_2 \Rightarrow B_1C_1=kc\) etc.
\[(\Large(\text( middle line triangle)))\]
Definition
The midline of a triangle is a segment connecting the midpoints of any two sides of the triangle.
Theorem
The middle line of the triangle is parallel to the third side and equal to half of it.
Proof
1) The parallelism of the middle line to the base follows from what was proven above lemmas.
2) Let us prove that \(MN=\dfrac12 AC\) .
Through the point \(N\) we draw a line parallel to \(AB\) . Let this line intersect the side \(AC\) at the point \(K\) . Then \(AMNK\) is a parallelogram ( \(AM\parallel NK, MN\parallel AK\) according to the previous point). So, \(MN=AK\) .
Because \(NK\parallel AB\) and \(N\) are the midpoint of \(BC\), then by Thales’ theorem \(K\) is the midpoint of \(AC\) . Therefore, \(MN=AK=KC=\dfrac12 AC\) .
Consequence
The middle line of the triangle cuts off from it a triangle similar to the given one with the coefficient \(\frac12\) .
Sometimes topics that are explained in school may not always be clear the first time. This is especially true for a subject like mathematics. But everything becomes much more complicated when this science begins to be divided into two parts: algebra and geometry.
Each student may have an ability in one of two directions, but especially in primary school It is important to understand the basis of both algebra and geometry. In geometry, one of the main topics is considered to be the section on triangles.
How to find the midline of a triangle? Let's figure it out.
Basic Concepts
To begin with, to figure out how to find the middle line of a triangle, it is important to understand what it is.
There are no restrictions on drawing the middle line: the triangle can be anything (isosceles, equilateral, rectangular). And all properties that relate to the middle line will be in effect.
The midline of a triangle is a segment connecting the midpoints of its 2 sides. Therefore, any triangle can have 3 such lines.
Properties
To know how to find the midline of a triangle, let’s designate its properties that need to be remembered, otherwise without them it will be impossible to solve problems with the need to designate the length of the midline, since all the data obtained must be substantiated and argued with theorems, axioms or properties.
Thus, to answer the question: “How to find the midline of triangle ABC?”, it is enough to know one of the sides of the triangle.
Let's give an example
Take a look at the picture. It shows triangle ABC with middle line DE. Note that it is parallel to the base AC in the triangle. Therefore, whatever the value of AC, the average line DE will be half as large. For example, AC=20 means DE=10, etc.
In these simple ways you can understand how to find the middle line of a triangle. Remember its basic properties and definition, and then you will never have problems finding its meaning.
1 Additional construction leading to the triangle midline theorem, trapezoid and similarity properties of triangles.
And she equal to half the hypotenuse.
Corollary 1.
Corollary 2.
From this it is clear that 
1 All right triangles with the same acute angle are similar. A look at trigonometric functions.
Triangles with hatched and non-hatched sides are similar in that their two angles are equal. Therefore where
This means that these relationships depend only on acute angle right triangle and essentially define it. This is one of the reasons for the appearance trigonometric functions:
Often writing trigonometric functions of angles in similar right triangles is clearer than writing similarity relations!
2 An example of an additional construction is a height lowered to the hypotenuse. Derivation of the Pythagorean theorem based on the similarity of triangles.
Let us lower the height CH to the hypotenuse AB. We have three similar to a triangle ABC, AHC and CHB. Let's write down expressions for trigonometric functions:
From this it is clear that . Adding up, we get the Pythagorean theorem, since:
For another proof of the Pythagorean theorem, see the commentary to Problem 4.
3 An important example of an additional construction is the construction of an angle equal to one of the angles of a triangle.
We carry out from the top right angle a straight segment making an angle with leg CA, equal to angle CAB of the given right triangle ABC. As a result, we obtain an isosceles triangle ACM with base angles. But the other triangle resulting from this construction will also be isosceles, since each of its angles at the base is equal (by the property of the angles of a right triangle and by construction - the angle was “subtracted” from the right angle). Due to the fact that triangles BMC and AMC are isosceles with common side MC, we have the equality MB=MA=MC, i.e. M.C. median drawn to the hypotenuse of a right triangle, and she equal to half the hypotenuse.
Corollary 1. The midpoint of the hypotenuse is the center of the circle circumscribed around this triangle, since it turns out that the midpoint of the hypotenuse is equidistant from the vertices of the right triangle.
Corollary 2. The middle line of a right triangle, connecting the middle of the hypotenuse and the middle of the leg, is parallel to the opposite leg and is equal to half of it.
In isosceles triangles BMC and AMC, let us lower the heights MH and MG to the bases. Since in isosceles triangle, the height lowered to the base is also the median (and bisector), then MH and MG are the lines of a right triangle connecting the middle of the hypotenuse with the midpoints of the legs. By construction, they turn out to be parallel to the opposite legs and equal to their halves, since the triangles are equal MHC and MGC are equal (and MHCG is a rectangle). This result is the basis for the proof of the theorem on the midline of an arbitrary triangle and, further, the midline of a trapezoid and the property of proportionality of segments cut off by parallel lines on two straight lines intersecting them.
Tasks
Using similarity properties -1
Using basic properties - 2
Using additional formation 3-4
1 2 3 4
The height dropped from the vertex of a right angle of a right triangle is equal to the square root of the lengths of the segments into which it divides the hypotenuse.
The solution seems obvious if you know the derivation of the Pythagorean theorem from the similarity of triangles:
\(\mathrm(tg)\beta=\frac(h)(c_1)=\frac(c_2)(h)\),
whence \(h^2=c_1c_2\).
Find the locus of points (GMT) of intersection of the medians of all possible right triangles whose hypotenuse AB is fixed.
The point of intersection of the medians of any triangle cuts off one third from the median, counting from the point of its intersection with the corresponding side. IN right triangle The median drawn from a right angle is equal to half the hypotenuse. Therefore, the desired GMT is a circle of radius equal to 1/6 of the length of the hypotenuse, with a center in the middle of this (fixed) hypotenuse.