Class: 9
Lesson objectives:
- generalize, expand and systematize students’ knowledge and skills; teach how to use knowledge when solving complex problems;
- promote the development of skills for independent application of knowledge when solving problems;
- develop students’ logical thinking and mathematical speech, the ability to analyze, compare and generalize;
- instill in students self-confidence and hard work; ability to work in a team.
Lesson objectives:
- Educational: repeat the theorems of Menelaus and Cheva; apply them when solving problems.
- Developmental: learn to put forward a hypothesis and skillfully defend your opinion with evidence; test your ability to generalize and systematize your knowledge.
- Educational: increase interest in the subject and prepare for solving more complex problems.
Lesson type: lesson of generalization and systematization of knowledge.
Equipment: cards for collective work in a lesson on this topic, individual cards for independent work, computer, multimedia projector, screen.
During the classes
Stage I. Organizational moment (1 min.)
The teacher announces the topic and purpose of the lesson.
Stage II. Updating basic knowledge and skills (10 min.)
Teacher: During the lesson, we will remember the theorems of Menelaus and Cheva in order to successfully move on to solving problems. Let's take a look at the screen where it is presented. For which theorem is this figure given? (Menelaus' theorem). Try to clearly formulate the theorem.
Picture 1
Let point A 1 lie on side BC of triangle ABC, point C 1 on side AB, point B 1 on the continuation of side AC beyond point C. Points A 1 , B 1 and C 1 lie on the same straight line if and only if equality holds 
Teacher: Let's look at the following picture together. State a theorem for this drawing.
Figure 2
Line AD intersects two sides and the extension of the third side of the IUD triangle.
According to Menelaus' theorem 
Straight line MB intersects two sides and the extension of the third side of triangle ADC.
According to Menelaus' theorem 
Teacher: What theorem does the picture correspond to? (Ceva's theorem). State the theorem.
Figure 3
Let point A 1 in triangle ABC lie on side BC, point B 1 on side AC, point C 1 on side AB. Segments AA 1, BB 1 and CC 1 intersect at one point if and only if the equality holds 
Stage III. Problem solving. (22 min.)
The class is divided into 3 teams, each receiving a card with two different tasks. Time is given to decide, then the following appears on the screen:<Рисунки 4-9>. Based on the completed drawings for the tasks, team representatives take turns explaining their solutions. Each explanation is followed by a discussion, answering questions, and checking the correctness of the solution on the screen. All team members take part in the discussion. The more active the team, the higher it is rated when summing up the results.
Card 1.
1. In triangle ABC, point N is taken on side BC so that NC = 3BN; on the continuation of side AC, point M is taken as point A so that MA = AC. Line MN intersects side AB at point F. Find the ratio
2. Prove that the medians of a triangle intersect at one point.
Solution 1
Figure 4
According to the conditions of the problem, MA = AC, NC = 3BN. Let MA = AC =b, BN = k, NC = 3k. Line MN intersects two sides of triangle ABC and the continuation of the third.
According to Menelaus' theorem 
Answer:
Evidence 2
Figure 5
Let AM 1, BM 2, CM 3 be the medians of triangle ABC. To prove that these segments intersect at one point, it is enough to show that 
Then, by Ceva's (converse) theorem, the segments AM 1, BM 2 and CM 3 intersect at one point.
We have: 
So, it has been proven that the medians of a triangle intersect at one point.
Card 2.
1. Point N is taken on the PQ side of the triangle PQR, and point L is taken on the PR side, and NQ = LR. The intersection point of the segments QL and NR divides QL in the ratio m:n, counting from point Q. Find
2. Prove that the bisectors of a triangle intersect at one point.
Solution 1
Figure 6
By condition, NQ = LR, Let NA = LR =a, QF = km, LF = kn. Line NR intersects two sides of triangle PQL and the continuation of the third.
According to Menelaus' theorem 
Answer:
Evidence 2
Figure 7
Let's show that 
Then, by Ceva's (converse) theorem, AL 1, BL 2, CL 3 intersect at one point. By the property of triangle bisectors
Multiplying the obtained equalities term by term, we obtain 
For the bisectors of a triangle, Cheva's equality is satisfied, therefore, they intersect at one point.
Card 3.
1. In triangle ABC, AD is the median, point O is the middle of the median. Straight line BO intersects side AC at point K. In what ratio does point K divide AC, counting from point A?
2. Prove that if a circle is inscribed in a triangle, then the segments connecting the vertices of the triangle with the points of contact of opposite sides intersect at one point.
Solution 1
Figure 8
Let BD = DC = a, AO = OD = m. Straight line BK intersects two sides and the extension of the third side of triangle ADC.
According to Menelaus' theorem 
Answer:
Evidence 2
Figure 9
Let A 1, B 1 and C 1 be the tangent points of the inscribed circle of triangle ABC. In order to prove that the segments AA 1, BB 1 and CC 1 intersect at one point, it is enough to show that Cheva’s equality holds: 
Using the property of tangents drawn to a circle from one point, we introduce the following notation: C 1 B = BA 1 = x, AC 1 = CB 1 = y, BA 1 = AC 1 = z.
Cheva's equality is satisfied, which means that the bisectors of the triangle intersect at one point.
Stage IV. Problem solving (independent work) (8 min.)
Teacher: The work of the teams is finished and now we will begin independent work on individual cards for 2 options.
Lesson materials for students’ independent work
Option 1. In a triangle ABC, the area of which is 6, on side AB there is a point K, dividing this side in the ratio AK:BK = 2:3, and on the side AC there is a point L, dividing AC in the ratio AL:LC = 5:3. The point Q of intersection of straight lines СК and BL is removed from straight line AB at a distance . Find the length of side AB. (Answer: 4.)
Option 2. On side AC in triangle ABC, point K is taken. AK = 1, KS = 3. On side AB, point L is taken. AL:LB = 2:3, Q is the point of intersection of straight lines BK and CL. Find the length of the altitude of triangle ABC dropped from vertex B. (Answer: 1.5.)
The work is submitted to the teacher for checking.
V stage. Lesson summary (2 min.)
Errors made are analyzed, original answers and comments are noted. The results of each team's work are summed up and grades are assigned.
Stage VI. Homework (1 min.)
Homework is made up of problems No. 11, 12 pp. 289-290, No. 10 p. 301.
Final words from the teacher (1 min).
Today you heard each other’s mathematical speech from the outside and assessed your capabilities. In the future, we will use such discussions for a greater understanding of the subject. Arguments in the lesson were friends with facts, and theory with practice. Thank you all.
Literature:
- Tkachuk V.V. Mathematics for applicants. – M.: MTsNMO, 2005.
Mathematics - 10th grade Mendel Viktor Vasilievich, Dean of the Faculty of Natural Sciences, Mathematics and Information Technologies of the Far East State University CHEVA'S THEOREM AND MENELAY'S THEOREM A special place in planimetry is given to two remarkable theorems: Ceva's theorem and Menelaus' theorem. These theorems are not included in the basic curriculum of high school geometry, but their study (and application) is recommended for anyone who is interested in mathematics a little more than is possible within the framework of the school curriculum. Why are these theorems interesting? First, we note that when solving geometric problems, two approaches are productively combined: - one is based on the definition of a basic structure (for example: a triangle - a circle; a triangle - a secant line; a triangle - three straight lines passing through its vertices and intersecting at one point; a quadrilateral with two parallel sides, etc.) - and the second is the method of support problems (simple geometric problems to which the process of solving a complex problem is reduced). So, the theorems of Menelaus and Cheva are among the most frequently encountered constructions: the first considers a triangle, the sides or extensions of the sides of which are intersected by some line (secant), the second deals with a triangle and three lines passing through its vertices, intersecting at one point. Menelaus's theorem This theorem shows the observable (together with the inverse) relationships of segments, a pattern connecting the vertices of a triangle and the intersection points of a secant with the sides (extensions of the sides) of the triangle. The drawings show two possible cases of the location of the triangle and the secant. In the first case, the secant intersects two sides of the triangle and the extension of the third, in the second - the continuation of all three sides of the triangle. Theorem 1. (Menelaus) Let ABC be intersected by a straight line that is not parallel to side AB and intersects its two sides AC and BC, respectively, at points B1 and A1, and straight line AB at point C1, then AB1 CA1 BC1 1. B1C A1B C1 A Theorem 2. (converse to Menelaus’ theorem) Let the points A1, B1, C1 in triangle ABC belong to the straight lines BC, AC, AB, respectively, then if AB1 CA1 BC1 1 B1C A1B C1 A, then the points A1, B1, C1 lie on one straight line. The proof of the first theorem can be carried out as follows: perpendiculars from all vertices of the triangle are lowered onto the secant line. The result is three pairs of similar right triangles. The relations of the segments appearing in the formulation of the theorem are replaced by the relations of perpendiculars corresponding to them in similarity. It turns out that each perpendicular segment in fractions will be present twice: once in one fraction in the numerator, a second time in another fraction in the denominator. Thus, the product of all these ratios will be equal to one. The converse theorem can be proved by contradiction. It is assumed that if the conditions of Theorem 2 are met, points A1, B1, C1 do not lie on the same straight line. Then straight line A1B1 will intersect side AB at point C2, different from point C1. In this case, by virtue of Theorem 1, for points A1, B1, C2 the same relation will hold as for points A1, B1, C1. It follows from this that points C1 and C2 will divide the segment AB in the same ratios. Then these points coincide - we get a contradiction. Let's look at examples of the application of Menelaus' theorem. Example 1. Prove that the medians of a triangle at the intersection point are divided in the ratio 2:1 starting from the vertex. Solution. Let us write down the relation obtained in the theorem, Menelaus for the triangle ABMb and the straight line McM(C): AM c BM M bC 1. M c B MM b CA The first fraction in this product is obviously equal to 1, and the third second ratio is equal to 1. Therefore 2 2:1, which is what needed to be proven. Example 2. A secant intersects the extension of side AC of triangle ABC at point B1 so that point C is the midpoint of segment AB1. This secant divides side AB in half. Find in what ratio it divides side BC? Solution. For a triangle and a secant, let us write the product of three ratios from Menelaus’s theorem: AB1 CA1 BC1 1. B1C A1B C1 A From the conditions of the problem it follows that the first ratio is equal to one, and the third is 1, 2, so the second ratio is equal to 2, i.e. That is, the secant divides side BC in a ratio of 2:1. We will see the next example of the application of Menelaus's theorem when we consider the proof of Ceva's theorem. Ceva's Theorem Most of the remarkable points of a triangle can be obtained using the following procedure. Let there be some rule according to which we can choose a certain point A1 on side BC (or its continuation) of triangle ABC (for example, choose the midpoint of this side). Then we will construct similar points B1, C1 on the other two sides of the triangle (in our example, two more midpoints of the sides). If the selection rule is successful, then the lines AA1, BB1, CC1 will intersect at some point Z (the choice of the midpoints of the sides in this sense, of course, is successful, since the medians of the triangle intersect at one point). I would like to have some general method that allows one to determine from the position of points on the sides of a triangle whether the corresponding triple of lines intersects at one point or not. The universal condition that “closed” this problem was found in 1678 by the Italian engineer Giovanni Ceva. Definition. Segments connecting the vertices of a triangle with points on opposite sides (or their extensions) are called cevians if they intersect at one point. There are two possible locations for the cevians. In one variant, the intersection point is internal, and the ends of the cevians lie on the sides of the triangle. In the second option, the intersection point is external, the end of one cevian lies on the side, and the ends of the other two cevians lie on the extensions of the sides (see drawings). Theorem 3. (Cheva's direct theorem) In an arbitrary triangle ABC, on the sides BC, CA, AB or their extensions, points A1, B1, C1 are taken, respectively, such that the lines AA1, BB1, CC1 intersect at some common point, then BA1 CB1 AC1 1 CA1 AB1 BC1 . Proof: There are several original proofs of Ceva's theorem; we will consider a proof based on a double application of Menelaus' theorem. Let us write the relation of Menelaus' theorem the first time for the triangle ABB1 and the secant CC1 (we denote the point of intersection of the Cevians as Z): AC1 BZ B1C 1, C1B ZB1 CA and the second time for the triangle B1BC and the secant AA1: B1Z BA1 CA 1. ZB A1C AB1 Multiplying these two ratios and making the necessary reductions, we obtain the ratio contained in the statement of the theorem. Theorem 4. (Ceva's converse theorem). If for the points A1, B1 and C1 selected on the sides of the triangle ABC or their extensions, Cheva’s condition is satisfied: BA1 CB1 AC1 1 CA1 AB1 BC1, then the lines AA1, BB1 and CC1 intersect at one point. The proof of this theorem is carried out by contradiction, just like the proof of Menelaus' theorem. Let us consider examples of the application of Ceva's direct and inverse theorems. Example 3. Prove that the medians of a triangle intersect at one point. Solution. Consider the relation AC1 BA1 CB1 C1B A1C B1 A for the vertices of the triangle and the midpoints of its sides. Obviously, in each fraction the numerator and denominator have equal segments, so all these fractions are equal to one. Consequently, Cheva's relation is satisfied, therefore, by the converse theorem, the medians intersect at one point. Problems for independent solution The problems proposed here are test work No. 1 for 9th grade students. Solve these problems, write down the solutions in a separate notebook (from physics and computer science). Indicate the following information about yourself on the cover: 1. Last name, first name, class, class profile (for example: Vasily Pupkin, 9th grade, mathematics) 2. Zip code, residence address, email (if any), telephone (home or mobile) ) 3. Information about the school (for example: MBOU No. 1, Bikin village) 4. Last name, full name of the mathematics teacher (for example: mathematics teacher Petrova M.I.) It is recommended to solve at least four problems. M 9.1.1. Can the secant line from Menelaus’ theorem cut the sides of a triangle (or their extensions) into segments of length: a) 3, 3, 5, 7,10, 14; c) 3, 5, 6, 7, 7, 10, If such options are possible, give examples. The segments can go in different orders. M 9.1.2. Can the interior cevians of a triangle divide its sides into segments: a) 3, 3, 5, 7,10, 14; c) 3, 5, 6, 7, 7, 10, If such options are possible, give examples. The segments can go in different orders. Hint: when coming up with examples, do not forget to check the triangle is not identical. M 9.1.3. Using Ceva's converse theorem, prove that: a) the bisectors of a triangle intersect at one point; b) the segments connecting the vertices of the triangle with points on opposite sides, at which these sides touch the inscribed circle, intersect at one point. Directions: a) remember in what ratio the bisector divides the opposite side; b) use the property that segments of two tangents drawn from one point to a certain circle are equal. M 9.1.4. Complete the proof of Menelaus' theorem begun in the first part of the article. M 9.1.5. Prove that the altitudes of a triangle intersect at one point using Ceva's converse theorem. M 9.1.6. Prove Simpson's theorem: from an arbitrary point M taken on a circle circumscribed around triangle ABC, perpendiculars are dropped onto the sides or extensions of the sides of the triangle, prove that the bases of these perpendiculars lie on the same straight line. Hint: Use the converse of Menelaus' theorem. Try to express the lengths of the segments used in the relationships in terms of the lengths of the perpendiculars drawn from their point M. It is also useful to recall the properties of the angles of an inscribed quadrilateral.
Menelaus' theorem or the complete quadrilateral theorem has been known since the times of Ancient Greece. It received its name in honor of its author, an ancient Greek mathematician and astronomer. Menelaus of Alexandria(around 100 AD). This theorem is very beautiful and simple, but, unfortunately, it is not given due attention in modern school courses. Meanwhile, in many cases it helps to solve quite complex geometric problems very easily and elegantly.
Theorem 1 (Menelaus' theorem). Let ∆ABC be intersected by a line not parallel to side AB and intersecting its two sides AC and BC, respectively, at points F and E, and line AB at point D (Fig. 1),
then A F FC * CE EB * BD DA = 1
Note. To easily remember this formula, you can use the following rule: move along the contour of the triangle from the vertex to the point of intersection with the line and from the point of intersection to the next vertex.
Proof. From vertices A, B, C of the triangle we draw, respectively, three parallel lines until they intersect with the secant line. We get three pairs of similar triangles (a sign of similarity at two angles). The following equalities follow from the similarity of triangles:
Now let’s multiply these resulting equalities:
The theorem has been proven.
To feel the beauty of this theorem, let’s try to solve the geometric problem proposed below in two different ways: using auxiliary construction and with the help Menelaus' theorem.
Task 1.
In ∆ABC, the bisector AD divides side BC in the ratio 2:1. In what ratio does the median CE divide this bisector?
Solution.
Using auxiliary construction:
Let S be the intersection point of the bisector AD and the median CE. Let's build ∆ASB to parallelogram ASBK. (Fig. 2)
Obviously, SE = EK, since the intersection point of the parallelogram bisects the diagonals. Let us now consider the triangles ∆CBK and ∆CDS. It is easy to see that they are similar (a sign of similarity in two angles: and as internal one-sided angles with parallel lines AD and KB and a secant CB). From the similarity of the triangle the following follows:
Using the condition, we get:
CB CD = CD + DB CD = CD + 2CD CB = 3CD CD = 3
Now notice that KB = AS, like the opposite sides of a parallelogram. Then
AS SD = KB SD = CB CD = 3
Using Menelaus' theorem.
Let's consider ∆ABD and apply Menelaus' theorem to it (the line passing through the points C, S, E is a secant line):
BE EA * AS SD * DC CB = 1
According to the conditions of the theorem, we have BE/EA = 1, since CE is the median, and DC/CB = 1/3, as we already calculated earlier.
1 * AS SD * 1 3 = 1
From here we get AS/SD = 3 At first glance, both solutions are quite compact and approximately equivalent. However, the idea of an additional construction for schoolchildren often turns out to be very complex and not at all obvious, whereas, knowing Menelaus’ theorem, he only needs to apply it correctly.
Let's consider another problem in which Menelaus' theorem works very elegantly.
Task 2.
On sides AB and BC ∆ABC points M and N are given, respectively, such that the following equalities hold:
AM MB = CN NA = 1 2
In what ratio does the intersection point S of segments BN and CM divide each of these segments (Fig. 3)?
Solution.
Let's consider ∆ABN. Let's apply Menelaus' theorem for this triangle (the line passing through the points M, S, C is a secant line)
AM MB * BC SN * CN CA = 1
From the problem conditions we have: AM MB = 1 2
NC CA = NC CN + NA = NC CN + 2NC = NC 3 NC = 1 3
Let's substitute these results and get:
1 2 * BS SN * 1 3 = 1
Hence BS/SN = 6. And, therefore, the point S of intersection of segments BN and CM divides segment BN in a ratio of 6: 1.
Let's consider ∆ACM. Let's apply Menelaus' theorem for this triangle (the line passing through the points N, S, B is a secant line):
AN NC * CS SM * MB BA = 1
From the problem conditions we have: AN NC = 2
MB BA = MB BM + MA = 2MA 2MA + MA = 2MB 3MA = 2 3
Let's substitute these results and get:
2 * CS SM * 2 3 = 1
Hence CS/SM = 3/4
And, therefore, the point S of intersection of segments BN and CM divides segment CM in the ratio 3: 4.
The converse theorem to Menelaus' theorem is also true. It often turns out to be even more useful. It works especially well in proof problems. Often, with its help, even Olympiad problems are solved beautifully, easily and quickly.
Theorem 2(Converse theorem of Menelaus). Let a triangle ABC be given and the points D, E, F belong to the lines BC, AC, AB, respectively (note that they can lie both on the sides of the triangle ABC and on their extensions) (Fig. 4).
Then, if AF FC * CE EB * BD DA = 1
then points D, E, F lie on the same line.
Proof. Let us prove the theorem by contradiction. Let us assume that the relation from the conditions of the theorem is satisfied, but point F does not lie on the line DE (Fig. 5).
Let's denote the point of intersection of lines DE and AB with the letter O. Now we apply Menelaus' theorem and get: AE EC * CD DB * BO OA = 1
But, on the other hand, the equality BF FA = BO OA
cannot be executed.
Therefore, the relation from the conditions of the theorem cannot be satisfied. We got a contradiction.
The theorem has been proven.
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The geometry course contains theorems that are not studied in sufficient detail at school, but which can be useful for solving the most complex problems of the Unified State Exam and Unified State Exam. These include, for example, Menelaus' theorem. Traditionally, it is studied in classes with in-depth study of mathematics in the 8th grade, and in the regular program (according to Atanasyan’s textbook), Menelaus’ theorem is included in the textbook for grades 10-11.
Meanwhile, the result of studying Internet resources that mention Menelaus’ theorem shows that it is usually formulated incompletely and therefore inaccurately, and all cases of its use, as well as the proof of the converse theorem, are not given. The purpose of this article is to understand what Menelaus’ theorem is, how and why it is used, and also to share the methodology for teaching this theorem in individual tutor lessons with students.
Let's consider a typical problem (Task No. 26, OGE), which appears on exams in many variants, differing only in the numbers in the condition.
The solution to the problem itself is simple - you can find it below. In this article, we are mainly interested in a slightly different point, which is often omitted and taken for granted, as obvious. But the obvious is what can be proven. And this can be proven in various ways - usually they are proven exclusively using similarity - but it can also be done using Menelaus’ theorem.
It follows from the condition that since the angles at the lower base of the trapezoid add up to 90°, then if you extend the sides, you will get a right triangle. Next, from the resulting intersection point of the extensions of the side sides, draw a segment that passes through the middle of the bases. Why does this segment pass through all these three points? Usually, solutions to the problem found on the Internet do not say a word about this. There is not even a reference to the four-point trapezoid theorem, let alone a proof of this statement. Meanwhile, it can be proven using Menelaus’ theorem, which is the condition for three points to belong to one line.
Formulations of Menelaus' theorem
It's time to formulate the theorem. It should be noted that in various textbooks and manuals there are quite different formulations of it, although the essence remains unchanged. In the textbook by Atanasyan et al. for grades 10-11, the following formulation of Menelaus’ theorem is given, let’s call it “vector”:
In the textbook “Geometry grades 10-11” by Aleksandrov et al., as well as in the textbook by the same authors “Geometry. 8th grade” provides a slightly different formulation of Menelaus’s theorem, and it is the same for both grades 10-11 and grade 8:
Three notes need to be made here.
Note 1. There are no problems on exams that need to be solved only using vectors, for which “minus one” is used. Therefore, for practical use, the most convenient formulation is one that is essentially a corollary of the theorem for segments (this is the second formulation, highlighted in bold letters). We will limit ourselves to this for further study of Menelaus’ theorem, since our goal is to learn how to apply it to solve problems.
Note 2. Despite the fact that all textbooks clearly stipulate the case when all three points A 1, B 1 and C 1 can lie on the extensions of the sides of the triangle (or on straight lines containing the sides of the triangle), on several tutoring sites on the Internet only the case is formulated when two points lie on two sides, and the third one lies on the continuation of the third side. This can hardly be justified by the fact that in exams only problems of the first type are encountered and problems cannot be encountered when all these points lie on extensions of three sides.
Note 3. The converse theorem, i.e. the condition for three points to lie on the same line is usually not considered at all, and some tutors even advise (???) to study only the direct theorem and not consider the inverse theorem. Meanwhile, the proof of the converse statement is quite instructive and allows you to prove statements similar to those given in the solution to Problem 1. The experience of proving the converse theorem will undoubtedly provide tangible benefits to the student when solving problems.
Drawings and patterns
In order to teach a student to see Menelaus’ theorem in problems and use it when making decisions, it is important to pay attention to the pictures and patterns in the writing of the theorem for a specific case. And since the theorem itself is in its “pure” form, i.e. without surrounding by other segments, sides of various figures is usually not found in problems, then it is more appropriate to show the theorem on specific problems. And if you show drawings as an explanation, then make them multivariate. In this case, highlight in one color (for example, red) the straight line that is formed by three points, and in blue - the segments of the triangle involved in writing Menelaus’ theorem. In this case, those elements that do not participate remain black:
At first glance, it may seem that the formulation of the theorem is quite complex and not always understandable; after all, it involves three fractions. Indeed, if the student does not have enough experience, then he can easily make a mistake in writing, and, as a result, solve the problem incorrectly. And this is where problems sometimes begin. The point is that textbooks usually don't focus on how to "work around" when writing a theorem. Nothing is said about the laws of recording the theorem itself. That's why some tutors even draw different arrows to indicate the order in which the formula should be written. And they ask students to strictly follow such guidelines. This is partly correct, but it is much more important to understand the essence of the theorem than to write it down purely mechanically, using the “bypass rule” and arrows.
In fact, it is only important to understand the logic of the “bypass”, and it is so precise that it is impossible to make a mistake in writing the formula. In both cases a) and b) we write the formula for triangle AMC.
First, we define for ourselves three points - the vertices of the triangle. For us these are points A, M, C. Then we determine the points lying on the intersecting line (red line), these are B, P, K. We start the “movement” from the vertex of the triangle, for example, from point C. From this point we “go "to the point that is formed by the intersection, for example, of side AC and the intersecting line - for us this is point K. We write in the numerator of the first fraction - SK. Then from point K we “go” to the remaining point on line AC - to point A. We write KA in the denominator of the first fraction. Since point A also belongs to line AM, we do the same with segments on line AM. And here again, we start from the vertex, then we “go” to a point on the intersecting line, after which we move to the vertex M. “Having found ourselves” on the line BC, we do the same with the segments on this line. From M we “go”, of course, to B, after which we return to C. This “detour” can be done either clockwise or counterclockwise. It is only important to understand the rule of traversal - from a vertex to a point on a line, and from a point on a line to another vertex. This is roughly how the rule for writing the product of fractions is usually explained. The result is: 
Please note that the entire “detour” is reflected in the recording and, for convenience, is shown with arrows.
However, the resulting record can be obtained without performing any “traversal”. After the points are written out - the vertices of the triangle (A, M, C) and the points - lying on the intersecting line (B, P, K), also write down triplets of letters denoting points lying on each of the three lines. In our cases, these are I) B, M, C; II) A, P, M and III) A, C, K. After this, the correct left side of the formula can be written without even looking at the drawing and in any order. It is enough for us to write true fractions from each three letters that obey the rule - conventionally, the “middle” letters are the points of the intersecting line (red). Conventionally, the “outer” letters are the points of the triangle’s vertices (blue). When writing a formula in this way, you only need to make sure that any “blue” letter (the vertex of the triangle) appears once in both the numerator and the denominator. For example. 
This method is especially useful for cases of type b), as well as for self-testing.
Menelaus's theorem. Proof
There are several different ways to prove Menelaus' theorem. Sometimes they prove it using the similarity of triangles, for which a segment parallel to AC is drawn from point M (as in this drawing). Others draw an additional line that is not parallel to the intersecting line, and then, using straight lines parallel to the intersecting line, they seem to “project” all the necessary segments onto this line and, using a generalization of Thales’s theorem (i.e., the theorem on proportional segments), derive the formula. However, perhaps the simplest method of proof is obtained by drawing a straight line from point M parallel to the intersecting one. Let us prove Menelaus' theorem in this way.
Given: Triangle ABC. Line PK intersects the sides of the triangle and the continuation of side MC at point B.
Prove that the equality holds: 
Proof. Let's draw the ray MM 1 parallel to BK. Let us write down the relations in which the segments that are included in the formula of Menelaus’s theorem participate. In one case, consider lines intersecting at point A, and in the other case, intersecting at point C. 
Let's multiply the left and right sides of these equations: 
The theorem has been proven.
The theorem is proved similarly for case b).
From point C we draw a segment CC 1 parallel to straight line BK. Let us write down the relations in which the segments that are included in the formula of Menelaus’s theorem participate. In one case, consider lines intersecting at point A, and in the other case, intersecting at point M. Since Thales’s theorem does not say anything about the location of segments on two intersecting lines, the segments can be located on opposite sides of point M. Therefore, 
The theorem has been proven.
Now let's prove the converse theorem.
Given: 
Prove that points B, P, K lie on the same line. 
Proof. Let the straight line BP intersect AC at some point K 2 that does not coincide with the point K. Since BP is a straight line containing the point K 2 , then the just proven Menelaus theorem is valid for it. So, let's write it down for her 
However, we have just proven that 
It follows that Points K and K 2 coincide, since they divide the side AC in the same ratio.
For case b) the theorem is proved in a similar way.
Solving problems using Menelaus' theorem
First, let's return to Problem 1 and solve it. Let's read it again. Let's make a drawing:
Given a trapezoid ABCD. ST - midline of the trapezoid, i.e. one of the given distances. Angles A and D add up to 90°. We extend the sides AB and CD and at their intersection we get point K. Connect point K with point N - the middle of BC. Now we prove that point P, which is the midpoint of the base AD, also belongs to the line KN. Let us consider triangles ABD and ACD sequentially. Two sides of each triangle are intersected by line KP. Suppose the straight line KN intersects the base AD at some point X. By Menelaus’ theorem: 
Since triangle AKD is right-angled, point P, which is the midpoint of the hypotenuse AD, is equidistant from A, D and K. Similarly, point N is equidistant from points B, C and K. 
Where does one base equal 36 and the other equal 2. 
Solution. Consider triangle BCD. It is crossed by the ray AX, where X is the point of intersection of this ray with the extension of side BC. According to Menelaus' theorem: 
Substituting (1) into (2) we get: 
Solution. Let us denote by the letters S 1 , S 2 , S 3 and S 4 the areas of the triangles AOB, AOM, BOK and the quadrilateral MOKC, respectively. 
Since BM is the median, then S ABM = S BMC.
This means S 1 + S 2 = S 3 + S 4.
Since we need to find the ratio of the areas S 1 and S 4, we divide both sides of the equation by S 4: 
Let's substitute these values into formula (1): 
From the triangle BMC with secant AK, according to Menelaus’ theorem, we have: 
From triangle AKC with secant BM, by Menelaus’ theorem we have: 
All the necessary relations are expressed through k and now you can substitute them into expression (2): 
The solution to this problem using Menelaus' theorem is discussed on the page.
Math tutor's note. The application of Menelaus' theorem in this problem is the very case when this method allows you to significantly save time on the exam. This task is offered in the demo version of the entrance exam to the Lyceum at the Higher School of Economics for the 9th grade (2019).
© Mathematics tutor in Moscow, Alexander Anatolyevich, 8-968-423-9589.
Decide for yourself
1) The task is simpler.
On the median BD of triangle ABC a point M is marked so that BM: MD = m: n. Line AM intersects side BC at point K.
Find the ratio BK:KC.
2) The task is more difficult.
The bisector of angle A of parallelogram ABCD intersects side BC at point P, and diagonal BD at point T. It is known that AB: AD = k (0 3) Task No. 26 OGE.
In triangle ABC, the bisector BE and the median AD are perpendicular and have the same length equal to 36. Find the sides of triangle ABC.
Math tutor hint. On the Internet one can find a solution to such a problem using additional construction and then either similarity or finding the areas, and only after that the sides of the triangle. Those. both of these methods require additional construction. However, solving such a problem using the bisector property and Menelaus’ theorem does not require any additional constructions. It is much simpler and more rational.
A.V. Shevkin
FMS No. 2007
Theorems of Cheva and Menelaus on the Unified State Exam
A detailed article “Around the theorems of Ceva and Menelaus” was published on our website in the ARTICLES section. It is addressed to mathematics teachers and high school students who are motivated to become proficient in mathematics. You can return to it if you want to understand the issue in more detail. In this note we will provide brief information from the mentioned article and analyze solutions to problems from the collection for preparing for the Unified State Exam 2016.
Ceva's theorem
Let a triangle be given ABC and on its sides AB, B.C. And A.C. points marked C 1 , A 1 And B 1 accordingly (Fig. 1).
a) If the segments AA 1 , BB 1 and CC 1 intersect at one point, then
b) If equality (1) is true, then the segments AA 1 , BB 1 and CC 1 intersect at one point.
Figure 1 shows the case when the segments AA 1 , BB 1 and CC 1 intersect at one point inside the triangle. This is the so-called interior point case. Ceva's theorem is also valid in the case of an external point, when one of the points A 1 , B 1 or WITH 1 belongs to the side of the triangle, and the other two belong to the extensions of the sides of the triangle. In this case, the intersection point of the segments AA 1 , BB 1 and CC 1 lies outside the triangle (Fig. 2).
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How to remember Cheva's equality?
Let us pay attention to the technique of remembering equality (1). The vertices of the triangle in each relation and the relations themselves are written in the direction of traversing the vertices of the triangle ABC, starting from point A. From point A let's go to the point B, we meet the point WITH 1, write the fraction 
. Further from the point IN let's go to the point WITH, we meet the point A 1, write the fraction 
. Finally, from the point WITH let's go to the point A, we meet the point IN 1, write the fraction 
. In the case of an external point, the order of writing fractions is preserved, although the two “division points” of the segment are outside their segments. In such cases, they say that the point divides the segment externally.
Note that any segment connecting a vertex of a triangle with any point on a line containing the opposite side of the triangle is called ceviana.
Let's consider several ways to prove statement a) of Ceva's theorem for the case of an interior point. To prove Ceva's theorem, you need to prove statement a) by any of the methods proposed below, and also prove statement b). The proof of statement b) is given after the first method of proving statement a). The proof of Ceva's theorem for the case of an external point is carried out similarly.
Proof of statement a) of Ceva's theorem using the proportional segment theorem
Let three cevians AA 1 , BB 1 and CC 1 intersect at a point Z inside the triangle ABC.
The idea of the proof is to replace the relations of segments from equality (1) with the relations of segments lying on the same line.
Through the point IN Let's draw a straight line parallel to the cevian SS 1 . Straight AA 1 intersects the constructed line at the point M, and the straight line passing through the point C and parallel AA 1 , - at point T. Through dots A And WITH let's draw straight lines parallel to the cevians BB 1 . They will cross the line VM at points N And R accordingly (Fig. 3).
P 
about the theorem on proportional segments we have:
,
And 
.
Then the equalities are true
.
In parallelograms ZСTM And ZCRB segments TM, СZ And BR equal as opposite sides of a parallelogram. Hence, 
and the equality is true
.
To prove statement b) we use the following statement. Rice. 3
Lemma 1. If points WITH 1 and WITH 2 divide the segment AB internally (or externally) in the same relation, counting from the same point, then these points coincide.
Let us prove the lemma for the case when the points WITH 1 and WITH 2 divide the segment AB internally in the same relation: 
.
Proof. From equality 
equalities follow 
And 
. The last of them is satisfied only under the condition that WITH 1 B And WITH 2 B are equal, i.e., provided that the points WITH 1 and WITH 2 match.
Proof of the lemma for the case when the points WITH 1 and WITH 2 divide the segment AB Externally it is carried out similarly.
Proof of statement b) of Ceva's theorem
Let now equality (1) be true. Let us prove that the segments AA 1 , BB 1 and CC 1 intersect at one point.
Let the Chevians AA 1 and BB 1 intersect at a point Z, draw a segment through this point CC 2 (WITH 2 lies on the segment AB). Then, based on statement a) we obtain the correct equality
.
(2)
AND 
From comparison of equalities (1) and (2) we conclude that 
, i.e. points WITH 1 and WITH 2 divide the segment AB in the same relation, counting from the same point. From Lemma 1 it follows that the points WITH 1 and WITH 2 match. This means that the segments AA 1 , BB 1 and CC 1 intersect at one point, which is what needed to be proven.
It can be proven that the procedure for writing equality (1) does not depend on from which point and in which direction the vertices of the triangle are traversed.
Exercise 1. Find the length of the segment AN in Figure 4, which shows the lengths of other segments.
Answer. 8.
Task 2. Chevians A.M.,
BN, CK intersect at one point inside the triangle ABC. Find an attitude 
, If 
,
. Rice. 4
Answer.
.
P 
We present the proof of Ceva's theorem from the article. The idea of the proof is to replace the relations of segments from equality (1) with the relations of segments lying on parallel lines.
Let straight AA 1 , BB 1 , CC 1 intersect at a point O inside the triangle ABC(Fig. 5). Through the top WITH triangle ABC let's draw a straight line parallel AB, and its points of intersection with the lines AA 1 , BB 1 we denote accordingly A 2 , B 2 .
From the similarity of two pairs of triangles C.B. 2 B 1 And ABB 1 , BAA 1 And C.A. 2 A 1, Fig. 5
we have equalities
,
.
(3)
From the similarity of triangles BC 1 O And B 2 CO, AWITH 1 O And A 2 CO we have equalities 
, from which it follows that
.
(4)
P 
Multiplying equalities (3) and (4), we obtain equality (1).
Statement a) of Ceva's theorem has been proven.
Let us consider the proof of statement a) of Ceva's theorem using areas for an interior point. It is presented in the book by A.G. Myakishev and relies on statements that we formulate in the form of tasks 3 And 4 .
Task 3. The ratio of the areas of two triangles with a common vertex and bases lying on the same line is equal to the ratio of the lengths of these bases. Prove this statement.
Task 4. Prove that if 
, That 
And 
. Rice. 6
Let the segments AA 1 , BB 1 and CC 1 intersect at a point Z(Fig. 6), then
,
.
(5)
AND 
from equalities (5) and the second statement of the task 4
follows that 
or 
. Similarly we obtain that 
And 
. Multiplying the last three equalities, we get:
,
i.e., equality (1) is true, which is what needed to be proven.
Statement a) of Ceva's theorem has been proven.
Task 15. Let the cevians intersect at one point inside the triangle and divide it into 6 triangles whose areas are equal S 1 , S 2 , S 3 , S 4 , S 5 , S 6 (Fig. 7). Prove that . Rice. 7
Task 6. Find the area S triangle CNZ(the areas of other triangles are shown in Figure 8).
Answer. 15.
Task 7. Find the area S triangle CNO, if the area of the triangle ANO equals 10 and 
,
(Fig. 9).
Answer. 30.
Task 8. Find the area S triangle CNO, if the area of the triangle AB.C. equal to 88 and ,
(Fig. 9).
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R 
decision. Since , we denote 
,
. Because
, then we denote 
,
. From Ceva's theorem it follows that 
, and then 
. If 
, That 
(Fig. 10). We have three unknown quantities ( x, y
And S), so to find S Let's make three equations.
Because 
, That 
= 88. Since 
, That 
, where 
. Because 
, That 
.
So, 
, where 
. Rice. 10
Task 9.
In a triangle ABC points K And L belong respectively to the parties AB
And BC.
,
.
P AL And CK. Area of a triangle PBC is equal to 1. Find the area of the triangle ABC.
Answer. 1,75.
T 
Menelaus' theorem
Let a triangle be given ABC and on its sides A.C. And CB points marked B 1 and A 1 accordingly, and on the continuation side AB point marked C 1 (Fig. 11).
a) If the points A 1 , B 1 and WITH 1 lie on the same straight line, then
.
(6)
b) If equality (7) is true, then the points A 1 , B 1 and WITH 1 lie on the same straight line. Rice. eleven
How to remember Menelaus' equality?
The technique for remembering equality (6) is the same as for equality (1). The vertices of the triangle in each relation and the relations themselves are written in the direction of traversing the vertices of the triangle ABC- from vertex to vertex, passing through division points (internal or external).
Task 10. Prove that writing equality (6) from any vertex of the triangle in any direction produces the same result.
To prove Menelaus' theorem, you need to prove statement a) by any of the methods proposed below, and also prove statement b). The proof of statement b) is given after the first method of proving statement a).
Proof of statement a) using the proportional segment theorem
Iway. a) The idea of the proof is to replace the ratios of the lengths of segments in equality (6) with the ratios of the lengths of segments lying on the same line.
Let the points A 1 , B 1 and WITH 1 lie on the same straight line. Through the point C let's make a direct l, parallel to the line A 1 B 1, it intersects the line AB at the point M(Fig. 12).
R 
is. 12
By the theorem on proportional segments we have: 
And 
.
Then the equalities are true 
.
Proof of statement b) of Menelaus' theorem
Now let equality (6) be true, let us prove that the points A 1 , B 1 and WITH 1 lie on the same straight line. Let straight AB And A 1 B 1 intersect at a point WITH 2 (Fig. 13).
Since the points A 1 B 1 and WITH 2 lie on the same straight line, then according to statement a) of Menelaus’ theorem
. (7)
From a comparison of equalities (6) and (7) we have 
, from which it follows that the equalities are true
,
,
.
The last equality is true only if 
, i.e. if the points WITH 1 and WITH 2 match.
Statement b) of Menelaus' theorem has been proven. Rice. 13
Proof of statement a) using similarity of triangles
The idea of the proof is to replace the ratios of the lengths of segments from equality (6) with the ratios of the lengths of segments lying on parallel lines.
Let the points A 1 , B 1 and WITH 1 lie on the same straight line. From points A, B And C let's draw perpendiculars AA 0 , BB 0 and SS 0 to this straight line (Fig. 14).
R 
is. 14
From the similarity of three pairs of triangles A.A. 0 B 1 And CC 0 B 1 , CC 0 A 1 And BB 0 A 1 , C 1 B 0 B And C 1 A 0 A(at two angles) we have the correct equalities
,
,
,
multiplying them, we get:
.
Statement a) of Menelaus' theorem has been proven.
Proof of statement a) using areas
The idea of the proof is to replace the ratio of the lengths of the segments from equality (7) with the ratios of the areas of triangles.
Let the points A 1 , B 1 and WITH 1 lie on the same straight line. Let's connect the dots C And C 1 . Let us denote the areas of the triangles S 1 , S 2 , S 3 , S 4 , S 5 (Fig. 15).
Then the equalities are true
,
,
. (8)
Multiplying equalities (8), we get:
Statement a) of Menelaus' theorem has been proven.
R 
is. 15
Just as Ceva's theorem remains valid if the point of intersection of the Cevians is outside the triangle, Menelaus' theorem remains valid if the secant intersects only the extensions of the sides of the triangle. In this case, we can talk about the intersection of the sides of the triangle at the external points.
Proof of statement a) for the case of external points
P 
the secant intersects the sides of the triangle ABC at external points, i.e. intersects the extensions of the sides AB,B.C. And A.C. at points C 1 , A 1 and B 1, respectively, and these points lie on the same straight line (Fig. 16).
By the theorem on proportional segments we have:
And .
Then the equalities are true
Statement a) of Menelaus' theorem has been proven. Rice. 16
Note that the above proof coincides with the proof of Menelaus' theorem for the case when the secant intersects two sides of the triangle at the interior points and one at the exterior.
The proof of statement b) of Menelaus' theorem for the case of external points is similar to the proof given above.
Z 
assignment11.
In a triangle ABC points A 1 , IN 1 lie respectively on the sides Sun And AWITH.
P- point of intersection of segments AA 1
And BB 1 .
,
. Find an attitude 
.
Solution. Let's denote 
,
,
,
(Fig. 17). According to Menelaus' theorem for a triangle B.C.IN 1 and secant PA 1 we write the correct equality:
,
whence it follows that
. Rice. 17
Answer.
.
Z 
assignment12
(MSU, correspondence preparatory courses).
In a triangle ABC,
whose area is 6, on the side AB point taken TO,
sharing this side in relation 
, and on the side AC- dot L,
dividing AC in a relationship 
. Dot P
line intersections SK And INL
away from the straight line AB at a distance of 1.5. Find the side length AB.
Solution. From points R And WITH let's drop the perpendiculars PR And CM directly AB. Let's denote 
,
,
,
(Fig. 18). According to Menelaus' theorem for a triangle A.K.C. and secant P.L. Let's write down the correct equality: 
, from where we get that 
,
. Rice. 18
From the similarity of triangles TOM.C. And TOR.P.(at two angles) we get that 
, from which it follows that 
.
Now, knowing the length of the height drawn to the side AB triangle ABC, and the area of this triangle, we calculate the length of the side: 
.
Answer. 4.
Z 
assignment13.
Three circles with centers A,IN,WITH,
whose radii are related as 
, touch each other externally at points X, Y, Z as shown in Figure 19. Segments AX And BY intersect at a point O.
In what respect, counting from the point B, line segment CZ divides a segment BY?
Solution. Let's denote 
,
,
(Fig. 19). Because 
, then according to statement b) of Ceva’s theorem the segments AX, BY And WITHZ intersect at one point - a point O. Then the segment CZ divides a segment BY in a relationship 
. Let's find this relationship. Rice. 19
According to Menelaus' theorem for a triangle B.C.Y. and secant OX we have: 
, from which it follows that 
.
Answer.
.
Task 14 (Unified State Exam 2016).
Points IN 1 and WITH AC And AB triangle ABC, and AB 1:B 1 WITH
=
= AC 1:WITH 1 B. Direct BB 1
And SS 1
intersect at a point ABOUT.
A 
) Prove that the line JSC bisects the side Sun.
AB 1 O.C. 1 to the area of the triangle ABC, if it is known that AB 1:B 1 WITH = 1:4.
Solution. a) Let it be a straight line A.O. crosses the side B.C. at the point A 1 (Fig. 20). By Ceva's theorem we have:
.
(9)
Because AB 1:B 1 WITH
= AC 1:WITH 1 B, then from equality (9) it follows that 
, that is C.A. 1 = A 1 B, which was what needed to be proven. Rice. 20
b) Let the area of the triangle AB 1 O equal to S. Because AB 1:B 1 WITH C.B. 1 O equals 4 S, and the area of the triangle AOC equals 5 S. Then the area of the triangle AOB is also equal to 5 S, since triangles AOB And AOC have a common ground A.O., and their vertices B And C equidistant from the line A.O.. Moreover, the area of the triangle AOC 1 is equal S, because AC 1:WITH 1 B = 1:4. Then the area of the triangle ABB 1 equals 6 S. Because AB 1:B 1 WITH= 1:4, then the area of the triangle C.B. 1 O equal to 24 S, and the area of the triangle ABC equal to 30 S. Now let's find the ratio of the area of the quadrilateral AB 1 O.C. 1 (2S) to the area of the triangle ABC (30S), it is equal to 1:15.
Answer. 1:15.
Task 15 (Unified State Exam 2016).
Points IN 1 and WITH 1 lie on the sides respectively AC And AB triangle ABC, and AB 1:B 1 WITH
=
= AC 1:WITH 1 B. Direct BB 1
And SS 1
intersect at a point ABOUT.
a) Prove that the line JSC bisects the side Sun.
b) Find the ratio of the area of the quadrilateral AB 1 O.C. 1 to the area of the triangle ABC, if it is known that AB 1:B 1 WITH = 1:3.
Answer. 1:10.
Z 
assignment 16 (USE-2016). On the segment BD point taken WITH. Bisector B.L. ABC with base Sun BLD with base BD.
a) Prove that the triangle DCL isosceles.
b) It is known that cos 
ABC DL, i.e. triangle BD point taken WITH. Bisector B.L. isosceles triangle ABC with base Sun is the lateral side of an isosceles triangle BLD with base BD.
a) Prove that the triangle DCL isosceles.
b) It is known that cos ABC= . In what respect is the straight line D.L. divides the side AB?
Answer. 4:21.
Literature
1. Smirnova I.M., Smirnov V.A. Wonderful points and lines of the triangle. M.: Mathematics, 2006, No. 17.
2. Myakishev A.G. Triangle geometry elements. (Series “Library “Mathematical Education””). M.: MTsNMO, 2002. - 32 p.
3. Geometry. Additional chapters for the 8th grade textbook: A textbook for students of schools and classes with in-depth study / L.S. Atanasyan, V.F. Butuzov, S.B. Kadomtsev et al. - M.: Vita-Press, 2005. - 208 p.
4. Erdniev P., Mantsaev N. Theorems of Cheva and Menelaus. M.: Kvant, 1990, No. 3, pp. 56–59.
5. Sharygin I.F. Theorems of Cheva and Menelaus. M.: Kvant, 1976, No. 11, pp. 22–30.
6. Vavilov V.V. Medians and midlines of a triangle. M.: Mathematics, 2006, No. 1.
7. Efremov Dm. New triangle geometry. Odessa, 1902. - 334 p.
8. Mathematics. 50 variants of typical test tasks / I.V. Yashchenko, M.A. Volkevich, I.R. Vysotsky and others; edited by I.V. Yashchenko. - M.: Publishing house "Exam", 2016. - 247 p.
