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 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.
Leading from the top right angle a straight line segment making an angle with leg CA equal to angle CAB of a 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.
Sometimes the topics that are explained at school may not always be clear the first time. This is especially true for a subject such as mathematics. But things become much more complicated when this science begins to be divided into two parts: algebra and geometry.
Each student may have the 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 for drawing the midline: the triangle can be any (isosceles, equilateral, right-angled). And all the properties that relate to the middle line will work.
The midline of a triangle is a line segment that connects the midpoints of 2 of its sides. Therefore, any triangle can have 3 such lines.
Properties
In order to know how to find the middle line of a triangle, we denote 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 middle line, since all the data obtained must be substantiated and argued by theorems, axioms or properties.
Thus, to answer the question: "How to find the midline of the triangle ABC?", It is enough to know one of the sides of the triangle.
Let's give an example
Take a look at the drawing. It represents triangle ABC with midline DE. Note that it is parallel to the base AC in the triangle. Therefore, whatever the value of AC, the middle line DE will be half as large. For example, AC=20 means DE=10, etc.
In such 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.
Lesson topic
Middle line of the triangle
Lesson Objectives
To consolidate the knowledge of schoolchildren about triangles;
To introduce students to such a concept as the middle line of a triangle;
To form students' knowledge about the properties of triangles;
Continue to teach children the use of the properties of figures in solving problems;
Develop logical thinking, perseverance and attention of students.
Lesson objectives
To form the knowledge of schoolchildren about the middle line of triangles;
Check students' knowledge on the topics covered about triangles;
Check students' ability to solve problems.
Develop students' interest in exact sciences;
Continue to develop the ability of students to express their thoughts and master the mathematical language;
Lesson plan
1. The middle line of the triangle. Basic concepts.
2. Median line of a triangle, theorems and properties.
3. Repetition of previously studied material.
4. The main lines of the triangle and their properties.
5. Interesting Facts from the field of mathematics.
6. Homework.
Middle line of the triangle
The midline of a triangle is the line segment that connects the midpoints of the two sides of the given triangle.
Each triangle has three middle lines that form another new triangle located inside.
The vertices of the newly formed triangle are at the midpoints of the sides of the given triangle.
In each triangle there is an opportunity to draw three middle lines.
Now let's take a closer look at this topic. Look at the drawing of the triangle above. Before you is a triangle ABC, on which the middle lines are drawn. Segments MN, MP and NP form another triangle MNP inside this triangle.
Properties of the midline of a triangle
Each midline of a triangle connecting the midpoints of its sides has the following properties:
1. The middle line of a triangle is parallel to its third side and equal to half of it.
Thus, we see that side AC is parallel to MN, which is half the size of side AC.
2. The middle lines of a triangle divide it into four equal triangles.
If we look at triangle ABC, we see that the midlines MN, MP and NP divided it into four equal triangles, and as a result, triangles MBN, PMN, NCP and AMP were formed.
3. The middle line of the triangle cuts off from the given triangle a similar one, the area of \u200b\u200bwhich is equal to one fourth of the original triangle.
So, for example, in triangle ABC the median line MP cuts off from this triangle, forming triangle AMP, the area of which is equal to one-fourth of triangle ABC.
triangles
In previous classes, you have already studied such a geometric figure as a triangle and you know what types of triangles are, how they differ and what properties they have.
The triangle is one of the simplest geometric shapes, which have three sides, three angles and their area is limited by three points and three segments that connect these points in pairs.
So we remembered the definition of a triangle, and now let's repeat everything you know about this figure by answering the questions:
4. What types of triangles have you already studied? List them.
5. Define each type of triangle.
6. What is the area of a triangle?
7. What is the sum of the angles of this geometric figure?
8. What types of triangles do you know? Name them.
9. What kind of triangles do you know by the type of equal sides?
10. Define the hypotenuse.
11. How many acute angles can there be in a triangle?
The main lines of the triangle
The main lines of a triangle are: median, bisector, height and median perpendicular.
Median
The median of a triangle is the line segment that connects the vertex of the triangle with the midpoint of the opposite side of the given triangle.
Triangle median properties
1. She divides the triangle into two others, equal in area;
2. All medians of this figure intersect at one point. This point divides them in a ratio of two to one, starting from the top, and is called the center of gravity of the triangle;
3. Medians divide this triangle into six equal ones.
Bisector
A ray that emerges from a vertex and, passing between the sides of an angle, bisects it, is called the bisector of that angle.
And if the segment of the bisector of an angle connects its vertex with a point that lies on the opposite side of the triangle, then it is called the bisector of the triangle.
Triangle bisector properties
1. The bisector of an angle is the locus of points that are equidistant from the sides of a given angle.
2. The bisector of the interior angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides of the triangle.
3. The center of a circle inscribed in a triangle is the intersection point of the bisectors of the given figure.
Height
The perpendicular, which is drawn from the top to the figure to the straight line, which is the opposite side of the triangle, is called its height.
Triangle height properties
1. The height drawn from the vertex of the right angle divides the triangle into two similar ones.
2. If the triangle is acute-angled, then its two heights cut off similar triangles from the given triangle.
Median perpendicular
The median perpendicular of a triangle is a line that passes through the midpoint of a segment that is perpendicular to this segment.
Properties of the perpendicular bisectors of a triangle
1. Any point of the perpendicular bisector to the segment is equidistant from its ends. In this case, the converse will also be true.
2. The point of intersection of the medial perpendiculars, which are drawn to the sides of the triangle, is the center of the circle, which is circumscribed about this triangle.
Interesting facts from the field of mathematics
Will it be news to you to know that they wanted to send François Vieta to the fire for deciphering the secret correspondence of the Spanish government, because they believed that only the devil could find out the cipher, and a person could not do it.
Do you know that the first person who suggested numbering chairs, rows and seats was René Descartes? The theater aristocrats even asked the king of France to give Descartes a reward for this, but, alas, the king refused, because he believed that giving awards to a philosopher was below his dignity.
Because of the students who could memorize the Pythagorean theorem but could not understand it, this theorem was called the "donkey bridge". This meant that the student was a "donkey" who could not cross the bridge. AT this case The bridge was considered the Pythagorean theorem.
Storytelling writers dedicated their works not only to mythical heroes, people and animals, but also to mathematical symbols. So, for example, the author of the famous Little Red Riding Hood wrote a fairy tale about the love of a compass and a ruler.
Homework
1. Three triangles are shown in front of you, give an answer, are the lines drawn in the triangles average?
2. How many middle lines can be built in one triangle?
3. Triangle ABC is given. Find the sides of the triangle ABC if its midlines have the following dimensions: OF = 5.5 cm, FN = 8 cm, ON = 7 cm.
\[(\Large(\text(Similar triangles)))\]
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 \(\frac12\) .
If parallel lines intersecting the sides of an angle cut off equal segments on one of its sides, then they cut off equal segments on its other side.
Proof. Let A 1, A 2, A 3 be the intersection points of parallel lines on one of the sides of the angle and A 2 lies between A 1 and A 3 (Fig. 1).
Let B 1 B 2 , B 3 - corresponding points intersection of these lines with the other side of the angle. Let us prove that if А 1 А 2 = A 2 A 3 , then В 1 В 2 = В 2 В 3 .
Let us draw a line EF through the point B 2 parallel to the line A 1 A 3 . By the property of a parallelogram A 1 A 2 \u003d FB 2, A 2 A 3 \u003d B 2 E.
And since A 1 A 2 \u003d A 2 A 3, then FB 2 \u003d B 2 E.
Triangles B 2 B 1 F and B 2 B 3 E are equal in the second criterion. They have B 2 F = B 2 E as proven. The angles at the vertex B 2 are equal as vertical, and the angles B 2 FB 1 and B 2 EB 3 are equal as internal crosses lying with parallel A 1 B 1 and A 3 B 3 and secant EF. From the equality of triangles follows the equality of the sides: B 1 B 2 \u003d B 2 B 3. The theorem has been proven.
Using the Thales theorem, the following theorem is established.
Theorem 2. The middle line of a triangle is parallel to the third side and equal to half of it.
The midline of a triangle is the line segment that connects the midpoints of its two sides. In Figure 2, segment ED is the midline of triangle ABC.
ED - midline of triangle ABC
Example 1 Divide this segment into four equal parts.
Decision. Let AB be a given segment (Fig. 3), which must be divided into 4 equal parts.
Dividing a segment into four equal parts
To do this, draw an arbitrary half-line a through the point A and plot four successively equal segments AC, CD, DE, EK on it.
Connect points B and K with a line segment. Let us draw lines through the remaining points C, D, E, parallel to the line VC, so that they intersect the segment AB.
According to the Thales theorem, the segment AB is divided into four equal parts.
Example 2 The diagonal of the rectangle is a. What is the perimeter of a quadrilateral whose vertices are the midpoints of the sides of the rectangle?
Decision. Let figure 4 correspond to the condition of the problem.
Then EF is the midline of triangle ABC and, therefore, by Theorem 2, $$ EF = \frac(1)(2)AC = \frac(a)(2) $$
Similarly $$ HG = \frac(1)(2)AC = \frac(a)(2) , EH = \frac(1)(2)BD = \frac(a)(2) , FG = \frac( 1)(2)BD = \frac(a)(2) $$ and, therefore, the perimeter of the quadrilateral EFGH is 2a.
Example 3 The sides of a triangle are 2 cm, 3 cm and 4 cm, and its vertices are the midpoints of the sides of another triangle. Find the perimeter of the big triangle.
Decision. Let figure 5 correspond to the condition of the problem.
The segments AB, BC, AC are the middle lines of the triangle DEF. Therefore, by Theorem 2 $$ AB = \frac(1)(2)EF\ \ ,\ \ BC = \frac(1)(2)DE\ \ ,\ \ AC = \frac(1)(2)DF $$ or $$ 2 = \frac(1)(2)EF\ \ ,\ \ 3 = \frac(1)(2)DE\ \ \ ,\ \ 4 = \frac(1)(2)DF $$ whence $$ EF = 4\ \ ,\ \ DE = 6\ \ ,\ \ DF = 8 $$ and hence the perimeter of triangle DEF is 18 cm.
Example 4 In a right triangle, straight lines are drawn through the midpoint of its hypotenuse, parallel to its legs. Find the perimeter of the resulting rectangle if the legs of the triangle are 10 cm and 8 cm.
Decision. AT triangle ABC(fig.6)
∠ A straight line, AB = 10 cm, AC = 8 cm, KD and MD are midlines of triangle ABC, whence $$ KD = \frac(1)(2)AC = 4 cm. \\ MD = \frac(1) (2) AB = 5 cm. $$ The perimeter of the rectangle K DMA is 18 cm.