The middle line of the triangle. Hello, friends! Today, the theoretical material is connected with the triangle. As part of the exam, there is a group of tasks that use the property of its middle line. And not only in problems with triangles, but also with trapezoids. There was one in which I suggested just remembering these facts, now in more detail ...
What is the middle line of a triangle and what are its properties?
Definition. The midline of a triangle is a line segment that connects the midpoints of the sides of the triangle.
It is clear that there are three middle lines in the triangle. Let's show them:
Without any proof, you have probably already noticed that all four triangles formed are equal. This is true, but we will talk more about this later.
Theorem. The midline of a triangle connecting the midpoints of two given sides is parallel to the third side and equal to half of it.
Proof:
1. Let's look at triangles BMN and BAC. By condition, we have BM=MA, BN=NC. We can write:
Therefore, the triangles are similar in terms of two proportional sides and the angle between them (the second sign of similarity). What follows from this? But the fact that:
On the basis of parallel lines MN||AC.
2. It also follows from the similarity of triangles that
That is, MN is two times less. Proven!
Let's solve a typical problem.
In triangle ABC, points M, N, K are the midpoints of sides AB, BC, AC. Find the perimeter of triangle ABC if MN=12, MK=10, KN=8.
Decision. Of course, the first thing to check is the existence of the triangle MNK (and hence the existence of the triangle ABC). The sum of the two smaller sides must be greater than the third side, we write 10+8>12. Execute, hence the triangle exists.
Let's build a sketch:
Thus the perimeter of the triangle ABC is 24+20+16=60.
*Now more about the triangles obtained when constructing all three middle lines. Their equality is easily proved. Look:
They are equal on three sides. Of course, other signs apply here as well. We get that
How is this property used in the tasks included in the exam? I would especially like to focus on problems in stereometry. There are types in which we are talking about a triangular prism.
For example, the plane is said to pass through the midpoints of the sides of the base and is parallel to the third edge of the base. Questions are raised about the change in the surface area of the prism, its volume, and others.
So. Knowing and understanding the information above, you will immediately determine that this plane cuts off one fourth of the specified prism from the base and solve the problem orally. Here with such tasks.
That's all! All the best!
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Sincerely, Alexander Krutitskikh.
Figure 1 shows two triangles. Triangle ABC is similar to triangle A1B1C1. And the adjacent sides are proportional, that is, AB is related to A1B1 in the same way that AC is related to A1C1. It is from these two conditions that the similarity of triangles follows.
Figure 2 shows lines a and b secant to c. This creates 8 corners. Angles 1 and 5 are corresponding, if the lines are parallel, then the corresponding angles are equal, and vice versa. 
In figure 3, M is the middle of AB and N is the middle of AC, BC is the base. The segment MN is called the midline of the triangle. The theorem itself says - The median line of a triangle is parallel to the base and equal to half of it.
In order to prove that MN is the midline of a triangle, we need the second similarity test for triangles and the parallelism test for lines.
Triangle AMN is similar to triangle ABC in the second way. In similar triangles, the corresponding angles are equal, angle 1 is equal to angle 2, and these angles are corresponding at the intersection of two lines of a secant, therefore, the lines are parallel, MN is parallel to BC. Angle A overall, AM/AB = AN/AC = ½
The similarity coefficient of these triangles is ½, which means that ½ = MN/BC, MN = ½ BC 
So we found the middle line of the triangle, and proved the theorem on the middle line of the triangle, if you still do not understand how to find the middle line, watch the video below.
The midline of a triangle is a line segment that connects the midpoints of 2 of its sides. Accordingly, each triangle has three median lines. Knowing the quality of the midline, as well as the lengths of the sides of the triangle and its angles, it is possible to find the length of the midline.
You will need
1. Let in a triangle ABC MN be the midline connecting the midpoints of the sides AB (point M) and AC (point N). By property, the midline of the triangle connecting the midpoints of 2 sides is parallel to the third side and equal to its half. This means that the midline MN will be parallel to the side BC and equal to BC/2. Consequently, to determine the length of the midline of the triangle, it is sufficient to know the length of the side of this particular third side.
2. Let us now know the sides whose midpoints are connected by the median line MN, that is, AB and AC, as well as the angle BAC between them. Because MN is the middle line, then AM = AB/2, and AN = AC/2. Then, according to the cosine theorem, objectively: MN ^ 2 = (AM ^ 2) + (AN ^ 2) -2 * AM * AN * cos (BAC) = (AB^2/4)+(AC^2/4)-AB*AC*cos(BAC)/2. From here, MN = sqrt((AB^2/4)+(AC^2/4)-AB*AC*cos(BAC)/2).
3. If the sides AB and AC are known, then the midline MN can be found by knowing the angle ABC or ACB. Let, say, the angle ABC be famous. Because, by the property of the midline, MN is parallel to BC, then the angles ABC and AMN are corresponding, and, consequently, ABC = AMN. Then by the law of cosines: AN^2 = AC^2/4 = (AM^2)+(MN^2)-2*AM*MN*cos(AMN). Consequently, the MN side can be found from the quadratic equation (MN^2)-AB*MN*cos(ABC)-(AC^2/4) = 0.
A square triangle is more correctly referred to as a right triangle. The relationships between the sides and angles of this geometric figure are considered in detail in the mathematical discipline of trigonometry.
You will need
1. Discover side rectangular triangle with support for the Pythagorean theorem. According to this theorem, the square of the hypotenuse is equal to the sum of the squares of the legs: c2 \u003d a2 + b2, where c is the hypotenuse triangle, a and b are its legs. In order to apply this equation, you need to know the length of any 2 sides of a rectangular triangle .
2. If the conditions specify the dimensions of the legs, find the length of the hypotenuse. To do this, with the support of a calculator, extract the square root of the sum of the legs, each of which is squared in advance.
3. Calculate the length of one of the legs, if the dimensions of the hypotenuse and the other leg are known. Using a calculator, take the square root of the difference between the squared hypotenuse and the driven leg, also squared.
4. If the hypotenuse and one of the acute angles adjacent to it are given in the problem, use the Bradys tables. They give the values of trigonometric functions for a large number of angles. Use the calculator with sine and cosine functions, as well as trigonometry theorems that describe the relationship between the sides and angles of a rectangular triangle .
5. Find the legs using the basic trigonometric functions: a = c*sin ?, b = c*cos ?, where a is the leg opposite the corner?, b is the leg adjacent to the corner?. Similarly, calculate the size of the sides triangle, if the hypotenuse and another acute angle are given: b = c*sin ?, a = c*cos ?, where b is the leg opposite to the angle?, and is the leg adjacent to the angle?
6. In the case when we lead the leg a and the acute angle adjacent to it?, do not forget that in a right triangle the sum of the acute angles is invariably equal to 90 °: ? +? = 90°. Find the value of the angle opposite to the leg a:? = 90° -?. Or use the trigonometric reduction formulas: sin ? = sin (90° -?) = cos?; tg? = tg (90° – ?) = ctg ? = 1/tan?.
7. If we keep the leg a and the acute angle opposite to it?, using Bradis tables, a calculator and trigonometric functions, calculate the hypotenuse using the formula: c=a*sin?, leg: b=a*tg?.
Related videos
1 Additional construction leading to the theorem on the midline of a triangle, trapezoid and similarity properties of triangles.
And she equal to half the hypotenuse.
Consequence 1.
Consequence 2.
From here it is clear that 
1 All right triangles with the same acute angle are similar. A look at trigonometric functions.
Triangles with primed and unprimed sides are similar in terms of the equality of their two angles. Therefore, where
This means that these relations depend only on the acute angle of the right triangle and, in fact, determine it. This is one of the reasons for the appearance of trigonometric functions:
Often the record of the trigonometric functions of the angle in similar right-angled triangles is clearer than the record of similarity relations!
2 An example of an additional construction is the height lowered to the hypotenuse. Derivation of the Pythagorean theorem based on the similarity of triangles.
Let us lower the altitude CH to the hypotenuse AB. We have three similar triangles ABC, AHC and CHB. Let's write expressions for trigonometric functions:
From here it is clear that . Adding, we get the Pythagorean theorem, because:
For another proof of the Pythagorean theorem, see the commentary to problem 4.
3 An important example of additional construction is the construction of an angle equal to one of the angles of a triangle.
We draw a straight line segment from the vertex of the right angle, making an angle with the leg CA equal to the angle CAB of the given right triangle ABC. As a result, we get an isosceles triangle ACM with angles at the base. But the other triangle resulting from such a construction will also be isosceles, since each of its angles at the base is equal (by the property of the angles of a right-angled triangle and by construction, the angle was "subtracted" from the right angle). Due to the fact that triangles BMC and AMC are isosceles with a common side MC, we have the equality MB=MA=MC, i.e. MC- median drawn to the hypotenuse of a right triangle, and she equal to half the hypotenuse.
Consequence 1. The midpoint of the hypotenuse is the center of the circle circumscribed around this triangle, since it turned out that the midpoint of the hypotenuse is equidistant from the vertices of the right triangle.
Consequence 2. The median line of a right triangle, connecting the midpoint of the hypotenuse and the midpoint of the leg, is parallel to the opposite leg and equal to half of it.
In isosceles triangles BMC and AMC, let's drop the heights MH and MG to the bases. Since in an isosceles triangle, the height dropped to the base is also the median (and bisector), then MH and MG are the lines of a right triangle connecting the midpoint 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 to MHC and MGC are equal (moreover, MHCG is a rectangle). This result is the basis for proving the theorem on the median line of an arbitrary triangle and, further, the median line of a trapezoid and the proportionality property of segments cut off by parallel lines on two lines intersecting them.
Tasks
Using Similarity Properties -1
Using Basic Properties - 2
Using Additional Build 3-4
1 2 3 4
The height dropped from the vertex of the 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 the intersection of the medians of all possible right-angled triangles, the hypotenuse AB of which 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 a right triangle, the median drawn from the right angle is 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.
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 very primary essence and the necessary basic knowledge about the perception of spatial forms by a person. One such problem is the problem of finding the midpoint of a triangle. Today, this task is considered as a teaching method for the development of the intellectual abilities of schoolchildren. In the ancient world, the knowledge of how to find the middle of a triangle was also applied in practice: in land management, in the manufacture of various mechanisms, etc. What is the essence of this geometric puzzle?
What is a 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, from which the name of this geometric figure comes. It is important to know how the lines connecting vertices with opposite sides are called: height, bisector and median.
Height - a line perpendicular to the side opposite the vertex from which it is drawn; bisector - divides the 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 middle of the segment, because it is the intersection point of the medians of the triangle that is its middle.
Find the midpoints of the sides of the triangle. Finding the midpoint of a segment is also a classic geometric problem, for which you need a compass and a ruler without divisions. We put the compass needle at the end point of the segment and draw a semicircle larger than half of the segment in the middle of the latter. 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 connect the two points of intersection of the circle with a straight line using a ruler. This line intersects the original segment exactly in its middle. Now, knowing how to find the midpoint of the segment, we do this with each side of the triangle. After finding all the midpoints of the sides of the triangle, you are ready to construct its own midpoint.
We build the middle of the triangle. Connecting the vertices of the triangle with the midpoints of their opposite sides with straight lines, we get three medians. This may surprise someone, 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 is 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 carries many such exciting "mysteries", the knowledge of which helps to comprehend the harmony of the surrounding world and the nature of more complex things.
\[(\Large(\text(Similar triangles)))\]
Definitions
Two triangles are said to be 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 similarity coefficient 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 the 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 \(\dfrac(AB)(A_1B_1) = \dfrac(AC)(A_1C_1) = \dfrac(BC)(B_1C_1) = k\). Denote by letters \(S\) and \(S_1\) the areas of these triangles, respectively.
Since \(\angle A = \angle A_1\) , then \(\dfrac(S)(S_1) = \dfrac(AB\cdot AC)(A_1B_1\cdot A_1C_1)\)(according to the theorem on the ratio of the areas of triangles having an equal angle).
As \(\dfrac(AB)(A_1B_1) = \dfrac(AC)(A_1C_1) = k\), then \(\dfrac(S)(S_1) = \dfrac(AB)(A_1B_1)\cdot\dfrac(AC)(A_1C_1) = k\cdot k = k^2\), which was to be proved.
\[(\Large(\text(Triangle Similarity Tests)))\]
Theorem (the first criterion for the 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 triangle sum theorem \(\angle C = 180^\circ - \angle A - \angle B = 180^\circ - \angle A_1 - \angle B_1 = \angle C_1\), 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 the 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 was to be proved.
Theorem (the second criterion for the similarity of triangles)
If two sides of one triangle are proportional to two sides of another triangle and the angles included between these sides are equal, then such 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's prove that the triangles \(ABC\) and \(A"B"C"\) are similar. Given the first triangle similarity criterion, it suffices to show that \(\angle B = \angle B"\) .
Consider a triangle \(ABC""\) , where \(\angle 1 = \angle A"\) , \(\angle 2 = \angle B"\) . Triangles \(ABC""\) and \(A"B"C"\) are similar in the first triangle similarity criterion, then \(\dfrac(AB)(A"B") = \dfrac(AC"")(A"C")\).
On the other hand, according to the condition \(\dfrac(AB)(A"B") = \dfrac(AC)(A"C")\). It follows from the last two equalities 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 (the third criterion for the similarity of triangles)
If three sides of one triangle are proportional to three sides of another triangle, then such triangles are similar.
Proof
Let the sides of 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 triangle similarity criterion, it suffices to prove that \(\angle BAC = \angle A"\) .
Consider a triangle \(ABC""\) , where \(\angle 1 = \angle A"\) , \(\angle 2 = \angle B"\) .
The triangles \(ABC""\) and \(A"B"C"\) are similar in the first triangle similarity criterion, 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 \(BC = BC""\) , \(CA = C""A\) .
The triangles \(ABC\) and \(ABC""\) are equal in three sides, therefore, \(\angle BAC = \angle 1 = \angle A"\).
\[(\Large(\text(Thales' theorem)))\]
Theorem
If on one side of the angle we mark segments equal to each other and draw parallel straight lines through their ends, then these straight lines will cut off segments equal to each other on the second side.
Proof
Let's prove first lemma: If in \(\triangle OBB_1\) a line \(a\parallel BB_1\) is drawn through the midpoint \(A\) of the side \(OB\) , then it will intersect the side \(OB_1\) also in the middle.
Draw \(l\parallel OB\) through the point \(B_1\) . Let \(l\cap a=K\) . Then \(ABB_1K\) is a parallelogram, hence \(B_1K=AB=OA\) 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 us proceed 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, by this lemma \(OA_1=A_1B_1\) . Let's prove that \(A_1B_1=B_1C_1\) . Draw a line through the point \(B_1\) \(d\parallel OC\) , and let \(d\cap a=D_1, d\cap c=D_2\) . Then \(ABB_1D_1, BCD_2B_1\) are parallelograms, hence \(D_1B_1=AB=BC=B_1D_2\) . Thus, \(\angle A_1B_1D_1=\angle C_1B_1D_2\) like vertical, \(\angle A_1D_1B_1=\angle C_1D_2B_1\) as lying crosswise, 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 proportional segments on the sides of the angle.
Proof
Let parallel lines \(p\parallel q\parallel r\parallel s\) split one of the lines into segments \(a, b, c, d\) . Then these lines should divide the second straight line into segments \(ka, kb, kc, kd\), respectively, where \(k\) is a certain number, the same coefficient of proportionality of the segments.
Let's draw a straight line \(p\parallel OD\) through the point \(A_1\) (\(ABB_2A_1\) 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, let us 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.
\[(\Large(\text(Middle line of the triangle)))\]
Definition
The midline of a triangle is a line segment that connects 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 midline to the base follows from the above lemmas.
2) We prove that \(MN=\dfrac12 AC\) .
Draw a line through the point \(N\) parallel to \(AB\) . Let this line intersect the side \(AC\) at the point \(K\) . Then \(AMNK\) is a parallelogram ( \(AM\parallel NK, MN\parallel AK\) on the previous point). So \(MN=AK\) .
Because \(NK\parallel AB\) and \(N\) is the midpoint of \(BC\) , then by the Thales theorem, \(K\) is the midpoint of \(AC\) . Therefore, \(MN=AK=KC=\dfrac12 AC\) .
Consequence
The middle line of the triangle cuts off a triangle similar to the given one with the coefficient \(\frac12\) .
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