Advertising on the portal

Median. Visual Guide (2019)

The median and altitude of a triangle is one of the most fascinating and interesting topics in geometry. The term "Median" means the line or line segment that connects the vertex of a triangle to its opposite side. In other words, the median is a line that runs from the middle of one side of a triangle to the opposite vertex of the same triangle. Since a triangle has only three vertices and three sides, it means there can only be three medians.

Properties of the median of a triangle

  1. All medians of a triangle intersect at one point and are separated by this point in a ratio of 2:1, counting from the vertex. Thus, if you draw all three medians in a triangle, then their intersection point will divide them into two parts. The part that is located closer to the vertex will be 2/3 of the entire line, and the part that is located closer to the side of the triangle will be 1/3 of the line. The medians intersect at one point.
  2. Three medians drawn in one triangle divide this triangle into 6 small triangles whose area will be equal.
  3. How bigger side triangle from which the median comes, the smaller the median. Conversely, the shortest side has the longest median.
  4. Median in right triangle has a number of its own characteristics. For example, if we describe a circle around such a triangle that will pass through all the vertices, then the median right angle drawn to the hypotenuse will become the radius of the circumscribed circle (that is, its length will be the distance from any point of the circle to its center).

Equation of the length of the median of a triangle

The median formula comes from Stewart's theorem and states that the median is square root from the ratio of the squares of the sum of the sides of the triangle that form the vertex, minus the square of the side to which the median is drawn to four. In other words, to find out the length of the median, you need to square the lengths of each side of the triangle, and then write it down as a fraction, the numerator of which will be the sum of the squares of the sides that form the angle from which the median comes, minus the square of the third side. The denominator here is the number 4. Then we need to extract the square root from this fraction, and then we will get the length of the median.

Intersection point of triangle medians

As we wrote above, all medians of one triangle intersect at one point. This point is called the center of the triangle. It divides each median into two parts, the length of which is proportional to 2:1. In this case, the center of the triangle is also the center of the circle circumscribed around it. And others geometric shapes have their own centers.

Coordinates of the point of intersection of the medians of the triangle

To find the coordinates of the intersection of the medians of one triangle, we use the property of the centroid, according to which it divides each median into segments 2:1. We denote the vertices as A(x 1 ;y 1), B(x 2 ;y 2), C(x 3 ;y 3),

and calculate the coordinates of the center of the triangle using the formula: x 0 = (x 1 + x 2 + x 3)/3; y 0 = (y 1 + y 2 + y 3)/3.

Area of ​​a triangle through the median

All medians of one triangle divide this triangle by 6 equal triangles, and the center of the triangle divides each median in a ratio of 2:1. Therefore, if the parameters of each median are known, you can calculate the area of ​​the triangle through the area of ​​one of the small triangles, and then increase this indicator by 6 times.

Proof. Let us prove that the medians AA 1 and CC 1 at the intersection point M are divided in the ratio 2:1. Theorem. The medians of a triangle intersect at one point and divide at this point in a ratio of 2: 1, counting from the vertices. Let D be the midpoint of segment BA 1. Then C 1 D is the midline of triangle ABA 1. Therefore, lines AA 1 and C 1 D are parallel. Since CA 1:A 1 D = 2:1, then by the theorem on proportional segments we obtain: CM:MC 1 = 2:1. Similarly, it is proved that the medians BB 1 and CC 1 at the intersection point are divided in the ratio 2:1. This means that all medians intersect at one point and are divided at this point in a ratio of 2: 1, counting from the vertices. Medians of a triangle


BC, then the median CM lies closer to the side AC, i.e. angle ACM is less than angle BCM. Exercise 1 Proof. Let us continue the median CM and set aside the segment MD equal to CM. Triangles AMD" title="Prove that if the inequality AC > BC holds for the sides of triangle ABC, then the median CM lies closer to the side AC, i.e. angle ACM is less than angle BCM. Exercise 1 Proof. Let's continue the median CM and plot the segment MD equal to CM. Triangles AMD." class="link_thumb"> 2 !} Prove that if the inequality AC > BC holds for the sides of triangle ABC, then the median CM lies closer to side AC, i.e. angle ACM is less than angle BCM. Exercise 1 Proof. Let us continue the median CM and set aside the segment MD equal to CM. Triangles AMD and BMC are equal in two sides and the angle between them. Therefore AD = BC. Since the smaller angle is opposite the smaller side of the triangle, angle ACD is less than angle ADC. This means that angle ACM is less than angle BCM. BC, then the median CM lies closer to the side AC, i.e. angle ACM is less than angle BCM. Exercise 1 Proof. Let us continue the median CM and set aside the segment MD equal to CM. Triangles AMD"> BC, then the median CM lies closer to side AC, i.e. angle ACM is less than angle BCM. Exercise 1 Proof. Let's continue the median CM and plot the segment MD equal to CM. Triangles AMD and BMC are equal in two sides and angle between them. Therefore, AD = BC. Since the smaller angle lies opposite the smaller side of the triangle, then the angle ACD is less than the angle ADC. This means that the angle ACM is less than the angle BCM."> BC, then the median CM lies closer to the side AC, i.e. . angle ACM is less than angle BCM. Exercise 1 Proof. Let us continue the median CM and set aside the segment MD equal to CM. Triangles AMD" title="Prove that if the inequality AC > BC holds for the sides of triangle ABC, then the median CM lies closer to the side AC, i.e. angle ACM is less than angle BCM. Exercise 1 Proof. Let's continue the median CM and plot the segment MD equal to CM. Triangles AMD."> title="Prove that if the inequality AC > BC holds for the sides of triangle ABC, then the median CM lies closer to side AC, i.e. angle ACM is less than angle BCM. Exercise 1 Proof. Let us continue the median CM and set aside the segment MD equal to CM. Triangles AMD"> !}


Prove that the median CM of triangle ABC is less than half the sum of sides AC and BC. Exercise 2 Proof. Let us continue the median CM and set aside the segment MD equal to CM. Triangles AMD and BMC are equal in two sides and the angle between them. Therefore AD = BC. By virtue of the triangle inequality, side CD less than the amount sides AC and AD. This means that the median CM of triangle ABC is less than half the sum of sides AC and BC.




The proof follows from the fact that the center of the circle circumscribed about a right triangle is the middle of the hypotenuse. Prove that the median of a right triangle drawn from the vertex of a right angle is equal to half the hypotenuse. Exercise 4


Let AB = c, AC = b, BC = a in triangle ABC. Prove that for the median m c drawn from vertex C, the formula holds. Proof. By the cosine theorem applied to triangles ACD and BCD, we have: Adding these equalities, we obtain an equality from which the required formula directly follows. Exercise 5










The area of ​​triangle ABC is 1. Find the area of ​​a triangle whose sides are equal to the medians of triangle ABC. Exercise 10 Solution. On the continuation of the segment MC 1, we plot the segment C 1 D equal to it. The sides of the triangle ADM are equal to two-thirds of the medians, and its area is equal to one-third. Therefore, the area of ​​a triangle whose sides are equal to the medians of triangle ABC is three-quarters. Answer. 0.75.


Theorem. The bisector of an angle of a triangle divides the opposite side into parts proportional to the adjacent sides. Bisectors of a triangle Proof. Let CD be the bisector of triangle ABC. Let us prove that AD: DB = AC: BC. Let us draw a straight line BE parallel to CD. In triangle BEC angle B equal to angle E. Therefore, BC = EC. By the proportional segment theorem, AD: DB = AC: CE = AC: BC.






Let AC = b, BC = a in triangle ABC. Prove that for the bisector l c drawn from vertex C, the formula holds where c, c are the segments into which the bisector divides side AB Proof. By the cosine theorem applied to triangles ACD and BCD, we have: Multiply the first equality by c, the second by c and add the resulting equalities. Doing identity transformations, we get equality. Exercise 3










Prove that the bisector of an angle of a triangle divides its area into parts proportional to the adjacent sides. Proof. Triangles AC 1 C and BC 1 C have a common altitude drawn from vertex C, and sides AC 1 and BC 1 are related as sides AC and BC. Therefore, the areas of triangles AC 1 C and BC 1 C are related as sides AC and BC. Exercise 8




Let AB = c, AC = b, BC = a in triangle ABC. Prove that the bisector CC 1 is divided by the intersection point of the bisectors in the ratio (a+b):c, counting from the vertex. Exercise 10 Proof. Let's draw a line C 1 C parallel to AA 1. Then A 1 C: CB = AC 1: C 1 B = b: a. Let A 1 C = bx, CB = ax. Since CA 1: A 1 B = b: c, then CA 1: A 1 C = b(a+b)x/c. Therefore, CO: OC 1 = (a + b)/c.


Altitudes of a triangle Theorem. In a right triangle, the perpendicular dropped from the right angle to the hypotenuse is the geometric mean of the projections of the legs onto the hypotenuse. (The geometric mean of two positive numbers a and b is called positive number c, whose square is equal to ab, i.e. c =). Proof. Triangles ADC and CDB are similar. Therefore, either CD 2 = AD BD, i.e. CD is the geometric mean of AD and BD.










Exercise 5 In triangle ABC, the altitudes AA 1 and BB 1 are drawn. Prove that the angles A 1 AC and B 1 BC are equal. Proof. A circle with diameter AB will pass through points A 1 and B 1. Inscribed angles A 1 AC and B 1 BC rest on one arc AB 1. Therefore, they are equal. To prove the equality of angles, one could use the fact that the sides of these angles are perpendicular.


Exercise 6 In triangle ABC, the altitudes AA 1 and BB 1 are drawn. Prove that the angles AA 1 B 1 and ABB 1 are equal. Proof. A circle with diameter AB will pass through points A 1 and B 1. Inscribed angles AA 1 B 1 and ABB 1 rest on the same arc AB 1. Therefore, they are equal.


Exercise 7 In triangle ABC, the altitudes AA 1 and BB 1 are drawn. Prove that the angles BAC and B 1 A 1 C are equal. Proof. Angle BAC is equal to 90 o minus angle ABB 1. Angle B 1 A 1 C is equal to 90 o minus angle AA 1 B 1. Since angles AA 1 B 1 and ABB 1 are equal (see the previous problem), then angles BAC and B 1 A 1 C.


Exercise 8 In triangle ABC, the altitudes AA 1 and BB 1 are drawn. Prove that triangle ABC is similar to triangle A 1 B 1 C. Proof. Angles BAC and B 1 A 1 C are equal (see previous problem). Angle C of triangles ABC and A 1 B 1 C is common. Therefore, these triangles are similar at two angles.






Exercise 11 Theorem. For the radius r of a circle inscribed in a triangle, the formula holds: where h a, h b, h c are the heights of the triangle. Proof. Let the sides of triangle ABC be a, b, c. For the area S of a triangle, the following equalities hold: From which the required formula follows.


Exercise 12 Prove that the points symmetrical to the point of intersection of the altitudes of a triangle relative to its sides lie on the circumcircle of this triangle. Proof. For point C, symmetrical point H of the intersection of the altitudes of triangle ABC, we have Therefore, point C belongs to the circumcircle. Similarly, the remaining two symmetrical points belong to the circumscribed circle.


Circle 1 Theorem 1. An angle with a vertex inside the circle is measured by the half-sum of the arcs on which this angle and the vertical angle with it rest. Proof. Consider the angle ACB with vertex C inside the circle and points A and B on the circle. Let A 1, B 1 be the points of intersection with the circle of the sides of the angle vertical to it. Let's draw the chord BB 1. Angle ACB is the external angle of triangle B 1 CB. Therefore, ACB = AB 1 B + B 1 BA 1. The angles on the right side of the equality are measured by the halves of the corresponding arcs, which completes the proof.


Circle 2 Theorem 2. The angle between the tangent to the circle and the chord drawn through the point of contact is measured by half the arc of the circle contained within this angle. Proof. Let the angle ACB be formed by the tangent AC and the chord BC of the circle. If this angle is a right angle, then BC is the diameter of the circle and, therefore, the angle ACB is measured by half the arc of the semicircle contained within this angle. If angle ACB is acute, then draw the diameter CD. We have ACB = ACD – BCD. Angle ACD is measured by half the arc CBD of a circle. Angle BCD is measured by half the arc BD of the circle. Therefore, their difference (angle ACB) is measured by half the arc CB of the circle contained within this angle. Consider the case of an obtuse angle yourself.


Circle 3 Theorem 3. An angle with a vertex outside the circle, the sides of which intersect the circle, is measured by the half-difference of the arcs of the circle contained within this angle. Proof. Consider an angle ACB with vertex C outside the circle and points A and B on the circle. Let A 1, B 1 be the points of intersection with the circle of sides AC and BC. Let's draw the chord AB 1. Angle AB 1 B is the exterior angle of triangle AB 1 C. Therefore, ACB = AB 1 B – B 1 AA 1. The angles on the right side of the equality are measured by halves of the corresponding arcs, which completes the proof.


Circle 4 Theorem 4. The product of segments of any chord drawn through an interior point of the circle is equal to the product of segments of a diameter drawn through the same point. Proof. Let us be given a circle with center at point O, chord AB and diameter CD intersect at point E. Let us prove that Triangles ACE and DBE are similar. Therefore, it means












































Exercise 21 Find the locus of the points from which a given segment AB is visible at a given angle, i.e. points C for which the angle ACB is equal this angle. Answer: Arcs of two circles of the same radius based on segment AB, without points A and B.




Exercise 23 Answer: a) HMTs lying outside the circle with diameter AB and not belonging to the straight line AB; For given points A and B, find the locus of points C for which the angle ACB is: a) acute; b) stupid. b) HMTs lying inside a circle with diameter AB and not belonging to the segment AB.






Exercise 26 Let AC and BD be chords of a circle intersecting at point E. Prove that triangles ABE and CDE are similar. Proof: Angle A of triangle ABE is equal to angle D of triangle CDE, as are inscribed angles subtended by the same arc of a circle. Similarly, angle B is equal to angle C. Therefore, triangles ABE and CDE are similar in the first respect.


Exercise 31 Answer: DEK and DLF, DEK and ELK, DLF and ELK, DFK and DLE, DFK and FLK, DLE and FLK. In the figure, DL is the bisector of triangle DEF inscribed in a circle. DL intersects the circle at point K, which is connected by segments to vertices E and F of the triangle. Find similar triangles.


Exercise 32 Answer: ABH and ADC, ACH and ADB, ABM and CDM, BMD and AMC. An acute triangle ABC is inscribed in a circle, AH is its height, AD is the diameter of the circle, which intersects side BC at point M. Point D is connected to vertices B and C of the triangle. Find similar triangles.


Exercise 33 Two straight lines are drawn through the external point E of the circle, intersecting the circle at points A, C and B, D, respectively. Prove that triangles ADE and BCE are similar. Proof: Angle D of triangle ADE is equal to angle C of triangle BCE, as are inscribed angles subtended by the same arc of a circle. The angle E of these triangles is common. Therefore, triangles ADE and BCE are similar in the first respect.


Exercise 34 Two straight lines are drawn through the external point E of the circle, intersecting the circle at points A, C and B, D, respectively. Prove that AE·CE = BE·DE. Proof: Triangles ADE and BCE are similar. So AE: DE = BE: CE. Therefore, AE·CE = BE·DE.
Exercise 36 A straight line is drawn through the outer point E of the circle, intersecting the circle at points A and B, and a tangent EC (C is the point of tangency). Prove that triangles EAC and ECB are similar. Proof. Triangles EAC and ECB share angle E. Angles ACE and CBE are equal, as are angles subtended by the same chord. Therefore, triangles EAC and ECB are similar.
78 A straight line touches circles of radii R and r at points A and B. It is known that the distance between the centers of the circles is equal to a, and r


A trapezoid with bases 14 and 40 is inscribed in a circle of radius 25. Find the height of the trapezoid. Solution. Let ABCD be a trapezoid inscribed in a circle with center O and radius 25. Two cases are possible: the bases AB and CD of the trapezoid are located on one side of the center O, the bases AB and CD are located on different sides from the center O. In the first case (Fig. 1), through point O we draw a line perpendicular to AB, and denote P, Q its points of intersection with AB and CD, respectively. Then the height PQ of the trapezoid is equal to OQ – OP. We have OQ = OP = Therefore, PQ = 9. In the second case (Fig. 2), through point O we draw a line perpendicular to AB, and denote P, Q its points of intersection with AB and CD, respectively. Then the height PQ of the trapezoid is equal to OQ + OP. We have OQ = OP = Therefore, PQ = 39. Answer. 9 or 39. Exercise 39


Circles with centers O 1 and O 2 intersect at points A and B. It is known that angle AO 1 B is equal to 90 o, angle AO 2 B is equal to 60 o, O 1 O 2 = a. Find the radii of the circles. Solution. Two cases are possible: points O 1, O 2 are located on opposite sides of line AB, points O 1, O 2 are located on one side of line AB. Let us denote by r the radius of the circle with center O 1. Then the radius of the circle with center O 2 will be equal. Let us denote by P the intersection point of the lines O 1 O 2 and AB. Then O 1 P =, O 2 P =. In the first case (Fig. 1) and, therefore, In the second case (Fig. 2) and, therefore, Answer. or Exercise 40


A circle with center O is circumscribed around triangle ABC and angle AOC is 60°. A circle with center M is inscribed in triangle ABC. Find angle AMC. In the first case (Fig. 1), the sum of angles A and C of triangle ABC is equal to 150°. Since AM and CM are bisectors of these angles, the sum of angles CAM and ACM is equal to 75° and, therefore, angle AMC is equal to 105°. Answer. 105 o or 165 o. Solution. There are two possible locations for vertex B of triangle ABC. In the second case (Fig. 2), the sum of angles A and C of triangle ABC is equal to 30°. Since AM and CM are bisectors of these angles, the sum of angles CAM and ACM is equal to 15 o and, therefore, angle AMC is equal to 165 o. Exercise 41


Triangle ABC is inscribed in a circle of radius 12. It is known that AB = 6 and BC = 4. Find AC. Solution. According to the theorem of sines, there are two possible cases of the location of vertex C of triangle ABC. Let us drop the perpendicular BH onto the line AC. Then BH = ABsinA = 1. By the Pythagorean theorem AH = CH = In the first case (Fig. 1) AC = In the second case (Fig. 2) AC = Answer. or Exercise 42


The lines containing the altitudes of triangle ABC intersect at point H. It is known that CH = AB. Find angle ACB. In the first case (Fig. 1), angle C is equal to angle CAA 1, like inscribed angles based on equal arcs. Therefore, angle C is 45°. In the second case (Fig. 2), angle C is 135°. Answer. 45 o or 135 o. Solution. Let AA 1, BB 1 be the altitudes of triangle ABC. Let us describe the circles on CH and AB as diameters. They will pass through points A 1 and B 1. There are two possible locations for point H. Exercise 43


In triangle ABC, the altitudes BB 1 and CC 1 are drawn, O is the center of the inscribed circle. It is known that BC = 24, B 1 C 1 = 12. Find the radius R of the circle circumscribed about the triangle BOC. Solution. There are two possible cases of location of the segment B 1 C 1. On BC, as on the diameter, we describe a circle with center P. The triangle B 1 C 1 P is equilateral. Therefore, the sum of angles BPB 1 and CPC 1 is 120 o. In the first case (Fig. 1), the triangles BPC 1 and CPB 1 are isosceles. Therefore, the sum of angles B and C is 120°. Since BO and CO are bisectors, the angle BOC is equal to 120°. Using the theorem of sines we find R =. In the second case (Fig. 2), the sum of angles B and C is equal to 60°. Since BO and CO are bisectors, the angle BOC is equal to 150°. Using the law of sines we find R = 24. Answer. or 24. Exercise 44

Median of a triangle- this is a segment connecting the vertex of a triangle with the middle of the opposite side of this triangle.

Properties of triangle medians

1. The median divides a triangle into two triangles of equal area.

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

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

Length of the median drawn to the side: ( proof by building up to a parallelogram and using the equality in a 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 a 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-vo: Let M be the intersection point of the medians CC 1, AA 1 of triangle ABC. Let's mark 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|| AC

and A 2 C 2 = 0.5*AC. WITH 1 A 1 - the middle line of triangle ABC. So A 1 WITH 1 || AC and A 1 WITH 1 = 0.5*AC.

Quadrangle A 2 C 1 A 1 C 2- a parallelogram, since its opposite sides are A 1 WITH 1 And A 2 C 2 equal and parallel. Hence, A 2 M = MA 1 And C 2 M = MC 1 . This means that the points A 2 And M divide the median AA 2 into three equal parts, i.e. AM = 2MA 2. Same as CM = 2MC 1 . So, point M of the intersection of two medians AA 2 And CC 2 triangle ABC divides each of them in the ratio 2:1, counting from the vertices of the triangle. It is proved in a completely similar way that the intersection point 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 point M, therefore, point M and there is the 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, - its median.

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

Triangle bisector. Theorems related to triangle bisectors. Formulas for finding bisectors

Angle bisector- a ray with a beginning at the vertex of an angle, dividing 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 interior 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).

Calculation of bisector length

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

a,b,c - sides of the triangle opposite vertices A,B,C, respectively,

p is the semi-perimeter of the triangle,

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

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

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


Area method.

Characteristics of the method. From the name it follows that the main object 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 elements of the figure (triangle). Therefore, a very effective technique is 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, by solving which we determine the unknown. This is where the main feature of the area method manifests itself - it “makes” an algebraic problem out of a geometric problem, 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 relation method: based on trace support problems:



Ceva's theorem

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

Proof.

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

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

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

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

Menelaus' theorem

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

Proof . Let us draw a line parallel to AB through point C. Let us 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 on side BC, and point B 1 on the continuation of side AC, and let the following relation hold:

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

A median is a segment drawn from the vertex of a triangle to the middle of the opposite side, that is, it divides it in half at the point of intersection. The point at which the median intersects the side opposite the vertex from which it emerges is called the base. Each median of the triangle passes through one point, called the intersection point. The formula for its length can be expressed in several ways.

Formulas for expressing the length of the median

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

where a, b and c are the sides. Moreover, c is the side on which the median falls. This is what it looks like simple formula. Medians of a triangle are sometimes required for auxiliary calculations. There are other formulas.

  • If, during the calculation, two sides of a 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 are divided by it in a ratio of two to one, if counted from the vertex. This point is called the center of gravity of the triangle.
  • The median divides the triangle into two others whose areas are equal. Such triangles are called equal-area.
  • If you draw all the medians, the triangle will be divided into 6 equal figures, which will also be triangles.
  • If all three sides of a triangle are equal, then each of the medians will also be an altitude and a bisector, that is, perpendicular to the side to which it is drawn, and bisects the angle from which it emerges.
  • IN isosceles triangle the median dropped from a vertex that is opposite a side that is not equal to any other will also be an altitude and a bisector. The medians dropped from other vertices are equal. This is also a necessary and sufficient condition for isosceles.
  • If the triangle is the base regular pyramid, then the height lowered to a given base is projected to the point of intersection of all medians.

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

  • The median of a triangle has another property. The formula for the square of its length through the squares of the sides is presented below.

Properties of the sides to which the median is drawn

  • If we connect any two intersection points of the medians with the sides on which they are dropped, then the resulting segment will be midline triangle and make up one half of the side of the triangle with which it does not have common points.
  • The bases of the altitudes and medians in a triangle, as well as the midpoints of the segments connecting the vertices of the triangle with the point of intersection of the altitudes, lie on the same circle.

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

There is a theorem that The medians of a triangle intersect at one point, and this point divides each median in a ratio of 2:1, where 2 corresponds to the segment from the vertex from which the median is drawn to the intersection point of the medians, and 1 corresponds to the segment from the intersection point of the medians to the middle of the side to which the median is drawn.

To prove this theorem, consider triangle ABC with medians AE, BF, CD. That is, points D, E, F bisect sides AB, BC, CA, respectively.
We do not know whether all medians intersect at one point (this still needs to be proven). However, any two medians will intersect at one point, since they cannot be parallel. Let medians AE and BF intersect at point O.

The median BF divides the median AE into two segments AO and EO. Let us draw a line parallel to BF through point E. This line will intersect side AC at a certain point L. We will also draw another line parallel to BF through the middle of segment AB (point D). It will intersect AC at point K.

According to Thales's theorem, if on one side of an angle from its vertex we lay out successively equal segments and draw parallel lines through the ends of these segments that intersect the other side of the angle, then these parallel lines will also cut off equal segments on the second side of the angle.

Let's look at angle BCA of this triangle. Segments BE and EC are equal to each other, lines BF and EL are parallel to each other. Then, according to Thales' theorem, CL = LF.
If we look at the angle BAC, since AD ​​= BD and DK || BF, then AK = KF.

Since the segments AF and CF are equal to each other (since they are formed by the median) and each of them is divided into two equal segments, then all four segments of the side AC are equal to each other: AK = KF = FL = LC.

Consider the angle EAC. Parallel lines are drawn through the ends of three equal segments of side AC. Consequently, they cut off equal segments on side AE. The segment AO contains two such segments, and EO only one. Thus, we have proven that at least one median of a triangle, at the point of intersection with another median, is divided into two segments, the lengths of which are related as 2: 1.

Now consider the intersection of the median AE with the median CD. Let them intersect at point P.

Similar to the previous one, it is proved that parallel lines FM, CD, EN divide side AB into equal segments. In turn, they divide AE ​​into three equal segments. Moreover, from vertex A to the intersection of the medians there are two such segments, and after that there is one.

One and the same segment cannot be divided into three equal parts so that with one division option they are the same size, and with another - a different one. Therefore, points O and P must coincide. This means that all three medians of the triangles intersect at one point.

To prove that the other two medians are divided by the intersection point in the ratio 2: 1, you can, similarly to the previous one, draw parallel lines to the sides AB and BC.