The medians of a triangle intersect at one point. Median

Triangle median is a line segment that connects the vertex of a triangle with the midpoint of the opposite side of this triangle.

Triangle median properties

1. The median divides the triangle into two triangles of the same area.

2. The medians of a triangle intersect at one point, which divides each of them in a ratio of 2:1, counting from the top. This point is called the center of gravity of the triangle (centroid).

3. The whole triangle is divided by its medians into six equal triangles.

The length of the median drawn to the side: ( doc by building up to a parallelogram and using the equality in the parallelogram of twice the sum of the squares of the sides and the sum of the squares of the diagonals )

T1. The three medians of the triangle intersect at one point M, which divides each of them in a ratio of 2:1, counting from the vertices of the triangle. Given: ∆ abc, SS 1, AA 1, BB 1 - medians
∆ABC. Prove: and

D-in: Let M be the intersection point of medians CC 1 , AA 1 of triangle ABC. Note A 2 - the middle of the segment AM and C 2 - the middle of the segment CM. Then A 2 C 2 is the middle line of the triangle AMS. Means, A 2 C 2|| AU

and A 2 C 2 \u003d 0.5 * AC. WITH 1 A 1 is the midline of triangle ABC. So A 1 WITH 1 || AC and A 1 WITH 1 \u003d 0.5 * AC.

quadrilateral A 2 C 1 A 1 C 2- a parallelogram, since its opposite sides A 1 WITH 1 And A 2 C 2 equal and parallel. Hence, A 2 M = MA 1 And C 2 M = MS 1 . This means that the points A 2 And M divide the median AA 2 into three equal parts, i.e. AM = 2MA 2. Similarly CM = 2MC 1 . So, the point M of the intersection of two medians AA 2 And CC2 triangle ABC divides each of them in the ratio 2:1, counting from the vertices of the triangle. Quite similarly, it is proved that the point of intersection of the medians AA 1 and BB 1 divides each of them in the ratio 2:1, counting from the vertices of the triangle.

On the median AA 1, such a point is the point M, therefore, the point M and there is a point of intersection of the medians AA 1 and BB 1.

Thus, n

T2. Prove that the segments that connect the centroid with the vertices of the triangle divide it into three equal parts. Given: ∆ABC , are its medians.

Prove: S AMB =S BMC =S-AMC.Proof. IN, they have in common. because their bases are equal and the height drawn from the top M, they have in common. Then

In a similar way, it is proved that S AMB = S AMC . Thus, S AMB = S AMC = S CMB .n

Bisector of a triangle. Theorems related to the bisectors of a triangle. Formulas for finding bisectors

Angle bisector A ray that starts at the vertex of an angle and divides the angle into two equal angles.

The bisector of an angle is the locus of points inside the angle that are equidistant from the sides of the angle.

Properties

1. Bisector theorem: The bisector of an interior angle of a triangle divides the opposite side in a ratio equal to the ratio of the two adjacent sides

2. The bisectors of the internal angles of a triangle intersect at one point - the incenter - the center of the circle inscribed in this triangle.

3. If two bisectors in a triangle are equal, then the triangle is isosceles (the Steiner-Lemus theorem).

Calculating the length of a bisector

l c - length of the bisector drawn to side c,

a,b,c - triangle sides against vertices A,B,C respectively,

p - half-perimeter of the triangle,

a l ,b l - lengths of the segments into which the bisector l c divides the side c,

α,β,γ - interior angles of the triangle at vertices A,B,C, respectively,

h c - the height of the triangle, lowered to side c.


area method.

Method characteristic. From the name it follows that the main object of this method is the area. For a number of figures, for example, for a triangle, the area is quite simply expressed through various combinations of the elements of the figure (triangle). Therefore, a technique is very effective when different expressions for the area of ​​a given figure are compared. In this case, an equation arises containing the known and desired elements of the figure, resolving which we determine the unknown. This is where the main feature of the area method manifests itself - from a geometric problem it “makes” an algebraic one, reducing everything to solving an equation (and sometimes a system of equations).

1) Comparison method: associated with a large number of formulas S of the same figures

2) S ratio method: based on the following reference tasks:



Ceva's theorem

Let the points A",B",C" lie on the lines BC,CA,AB of the triangle. The lines AA",BB",CC" intersect at one point if and only if

Proof.

Denote by the point of intersection of the segments and . Let us drop the perpendiculars from points C and A to the line BB 1 until they intersect with it at points K and L, respectively (see figure).

Since the triangles and have a common side, their areas are related as the heights drawn to this side, i.e. AL and CK:

The last equality is true, since right triangles and are similar in acute angle.

Similarly, we get And

Let's multiply these three equalities:

Q.E.D.

Comment. The segment (or continuation of the segment) connecting the vertex of the triangle with a point lying on the opposite side or its continuation is called ceviana.

Theorem (inverse Ceva theorem). Let the points A",B",C" lie on the sides BC,CA and AB of the triangle ABC respectively. Let the relation hold

Then the segments AA", BB", CC" and intersect at one point.

Theorem of Menelaus

Theorem of Menelaus. Let a line intersect triangle ABC, where C 1 is the point of its intersection with side AB, A 1 is its point of intersection with side BC, and B 1 is its point of intersection with the extension of side AC. Then

Proof . Draw a line through point C parallel to AB. Denote by K its point of intersection with the line B 1 C 1 .

Triangles AC 1 B 1 and CKB 1 are similar (∟C 1 AB 1 = ∟KCB 1 , ∟AC 1 B 1 = ∟CKB 1). Hence,

Triangles BC 1 A 1 and CKA 1 are also similar (∟BA 1 C 1 =∟KA 1 C, ∟BC 1 A 1 =∟CKA 1). Means,

From each equality we express CK:

Where Q.E.D.

Theorem (the inverse theorem of Menelaus). Let triangle ABC be given. Let point C 1 lie on side AB, point A 1 lie on side BC, and point B 1 lie on the extension of side AC, and the relation

Then the points A 1 ,B 1 and C 1 lie on the same straight line.

The median is the segment drawn from the vertex of the triangle to the middle of the opposite side, that is, it divides it in half by the point of intersection. The point at which the median intersects the opposite side from which it comes out is called the base. Through one point, called the point of intersection, passes each median of the triangle. The formula for its length can be expressed in several ways.

Formulas for expressing the length of the median

  • Often in problems in geometry, students have to deal with such a segment as the median of a triangle. The formula for its length is expressed in terms of the sides:

where a, b and c are sides. Moreover, c is the side on which the median falls. This is how the simplest formula looks like. Triangle medians are sometimes required for auxiliary calculations. There are other formulas as well.

  • If during the calculation two sides of the triangle and a certain angle α located between them are known, then the length of the median of the triangle, lowered to the third side, will be expressed as follows.

Basic properties

  • All medians have one common point of intersection O and they are also divided by it in a ratio of two to one, if we count from the top. This point is called the center of gravity of the triangle.
  • The median divides the triangle into two others, the areas of which are equal. Such triangles are called equal triangles.
  • If you draw all the medians, then the triangle will be divided into 6 equal figures, which will also be triangles.
  • If in a triangle all three sides are equal, then in it each of the medians will also be a height and a bisector, that is, perpendicular to the side to which it is drawn, and bisects the angle from which it exits.
  • In an isosceles triangle, the median dropped from a vertex that is opposite a side that is not equal to any other will also be the height and the bisector. Medians dropped from other vertices are equal. This is also a necessary and sufficient condition for isosceles.
  • If the triangle is the base of a regular pyramid, then the height lowered onto this base is projected to the intersection point of all medians.

  • In a right triangle, the median drawn to the longest side is half its length.
  • Let O be the point of intersection of the medians of the triangle. The formula below will be true for any point M.

  • Another property is the median of a triangle. The formula for the square of its length in terms of the squares of the sides is presented below.

Properties of the sides to which the median is drawn

  • If you connect any two points of intersection of the medians with the sides on which they are lowered, then the resulting segment will be the midline of the triangle and be one half from the side of the triangle with which it has no common points.
  • The bases of the heights and medians in the triangle, as well as the midpoints of the segments connecting the vertices of the triangle with the point of intersection of the heights, lie on the same circle.

In conclusion, it is logical to say that one of the most important segments is precisely the median of the triangle. Its formula can be used to find the lengths of its other sides.

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Lesson 1

Medians of a triangle. The point of intersection of the medians.

median A triangle is a line segment that connects the vertex of the triangle with the midpoint of the opposite side.

Proof:

The point of intersection of the medians of a triangle is center of gravity this triangle.

Task 1 The point of intersection of the medians of a triangle is separated from its vertices by distances equal to 4, 6 and 8. Find the lengths of the medians of the triangle.

Solution. Let AM, BE and CD be the medians in the triangle ABC, K is the point of their intersection, KC=4, KA=6 and KB=8.

https://pandia.ru/text/78/182/images/image004_34.gif" width="76" height="50">, that is, there are 2 parts for the KA segment, and one part for the KM segment, then the entire AM median consists of three equal parts and https://pandia.ru/text/78/182/images/image006_24.gif" width="104" height="41">.

Likewise,

,

Answer: 6, 9 and 12

Task 2 Medians AM and CK of triangle ABC are mutually perpendicular and equal to 6 and 9 respectively. Calculate the lengths of the sides AB and BC.

https://pandia.ru/text/78/182/images/image010_15.gif" width="104" height="41">,

That's why And

, .

Besides

, .

Using the Pythagorean theorem, we calculate the lengths of the segments AK and SM, we obtain

Now we calculate the lengths of the sides AB and BC:

AB=2AK=10, BC=2CM=.

Answer: 10;.

Test for self-control.

1. The median of a triangle divides in half (choose one of the answers)

1) triangle angle

2) the side of the triangle

3) two sides of a triangle

2. In what ratio does the intersection point of the medians of the triangle divide each of the medians of the triangle (choose the correct answers).

1) 2:1 counting from the base of the triangle

2) 1:2 counting from the top of the triangle

3) 2:1 counting from the top of the triangle

4) 1:2 counting from the base of the triangle

5) into two equal parts

3. If the median AM and P is the intersection point of the medians of the triangle in the triangle ABC, then what part of the median AM is the segment AP? (choose one of the answers)

4. In the triangle ABC, the median AM and P are drawn - the point of intersection of the medians of the triangle. What part of the median AM is the segment PM? (choose one of the answers)

5. In the triangle ABC, the median AM and P are drawn - the point of intersection of the medians of the triangle. What part of segment AP is segment PM? (choose one of the answers)

View the correct answers.

Tasks for independent solution.

1. The point of intersection of the medians of a triangle is separated from its vertices by distances equal to 6 cm, 8 cm and 12 cm. Find the lengths of the medians of the triangle.

View solution.

2. The medians VM and SC of the triangle ABC are mutually perpendicular and equal to 15 and 36 respectively. Find the lengths of the sides AB and AC.

View solution.

3. The medians of the triangle are 6, 9 and 12. How far from the vertices is the intersection point of the medians of the triangle?

View solution.

4. The medians of the triangle are 9, 12 and 18. Find the distances from the midpoints of the sides of the triangle to the center of gravity of this triangle.

View solution.

5. The center of gravity of a triangle is separated from the midpoints of its sides by distances. Equal to 5, 6 and 7. Find the medians of this triangle.

View solution.

6. The point of intersection of the medians of a triangle is removed from the midpoints of its sides by distances equal to 2, 3 and 4. At what distances from the vertices of the triangle is this point?

View solution.

Your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please read our privacy policy and let us know if you have any questions.

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  • When you submit an application on the site, we may collect various information, including your name, phone number, email address, etc.

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  • From time to time, we may use your personal information to send you important notices and communications.
  • We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
  • If you enter a prize draw, contest or similar incentive, we may use the information you provide to administer such programs.

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We do not disclose information received from you to third parties.

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  • In the event that it is necessary - in accordance with the law, judicial order, in legal proceedings, and / or based on public requests or requests from state bodies in the territory of the Russian Federation - disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public interest reasons.
  • In the event of a reorganization, merger or sale, we may transfer the personal information we collect to the relevant third party successor.

Protection of personal information

We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as from unauthorized access, disclosure, alteration and destruction.

Maintaining your privacy at the company level

To ensure that your personal information is secure, we communicate privacy and security practices to our employees and strictly enforce privacy practices.

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