THEOREMS OF CHEVA AND MENELAUS
Ceva's theorem
Most of the remarkable triangle points can be obtained using the following procedure. Let there be some rule according to which we can choose a certain point A 1 , on side BC (or its extension) of triangle ABC (for example, choose the midpoint of this side). Then we will construct similar points B 1, C 1 on the other two sides of the triangle (in our example there are two more midpoints of the sides). If the selection rule is successful, then straight AA 1, BB 1, CC 1 will intersect at some point Z (the choice of the midpoints of the sides in this sense is, of course, 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.
A universal condition that “closed” this problem was found in 1678 by an Italian engineerGiovanni Cheva .
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 version, the point
intersections are 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. (Ceva's direct theorem) In an arbitrary triangle ABC, points A are taken on sides BC, CA, AB or their extensions, respectively 1 , IN 1 , WITH 1 , such that straight AA 1 , BB 1 , SS 1 intersect at some common point, then
.
Proof: While several original proofs of Ceva's theorem are known, we will consider a proof based on a double application of Menelaus' theorem. Let us write down the relation of Menelaus’ theorem for the first time for a triangleABB 1 and secant CC 1 (we denote the point of intersection of the ceviansZ):
,
and the second time for a triangleB 1 B.C. and secant A.A. 1 :
.
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 those selected on the sides of the triangle ABC or their extensions of points A 1 , IN 1 And C 1 Cheva's condition is satisfied:
,
then straight A.A. 1 , BB 1 And CC 1 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
for the vertices of a 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.
Theorem (Ceva's theorem) . Let the points lie on the sides and triangle respectively. Let the segments And intersect at one point. Then
(we go around the triangle clockwise).
Proof. Let us denote by point of intersection of segments And . Let's omit from the points And perpendiculars to a linebefore intersecting it at points And accordingly (see figure).
Because triangles And have a common side, then their areas are related to the heights drawn to this side, i.e. And :
The last equality is true, since right triangles And similar in acute angle.
Similarly we get
And 
Let's multiply these three equalities:
Q.E.D.
About medians:
1. Place unit masses at the vertices of triangle ABC.
2. The center of mass of points A and B is in the middle of AB. The center of mass of the entire system must be at the median to side AB, since the center of mass of triangle ABC is the center of mass of the center of mass of points A and B, and point C.
(it got confusing)
3. Similarly - the CM must lie on the median to the sides AC and BC
4. Since the CM is a single point, then, therefore, all these three medians must intersect at it.
By the way, it immediately follows that by intersection they are divided in a ratio of 2:1. Since the mass of the center of mass of points A and B is 2, and the mass of point C is 1, therefore, the common center of mass, according to the proportion theorem, will divide the median in the ratio of 2/1.
Thank you very much, it is presented in an accessible way, I think it would not be amiss to present the proof using the methods of mass geometry, for example:
Lines AA1 and CC1 intersect at point O; AC1: C1B = p and BA1: A1C = q. We need to prove that line BB1 passes through point O if and only if CB1: B1A = 1: pq.
Let us place masses 1, p and pq at points A, B and C, respectively. Then point C1 is the center of mass of points A and B, and point A1 is the center of mass of points B and C. Therefore, the center of mass of points A, B and C with these masses is the intersection point O of lines CC1 and AA1. On the other hand, point O lies on the segment connecting point B with the center of mass of points A and C. If B1 is the center of mass of points A and C with masses 1 and pq, then AB1: B1C = pq: 1. It remains to note that on segment AC there is a single point dividing it in the given ratio AB1: B1C.
2. Ceva's theorem
A segment connecting a vertex of a triangle with a point on the opposite side is calledceviana . Thus, if in a triangleABC X , Y and Z - points lying on the sidesB.C. , C.A. , AB accordingly, then the segmentsAX , BY , CZ are Chevians. The term comes from the Italian mathematician Giovanni Ceva, who in 1678 published the following very useful theorem:
Theorem 1.21. If three cevians AX, BY, CZ (one from each vertex) of triangle ABC are competitive, then
|BX||XC|· |CY||YA|· |AZ||ZB|=1 .
| Rice. 3. |
When we say that three lines (or segments)competitive , then we mean that they all pass through one point, which we denote byP . To prove Ceva's theorem, recall that the areas of triangles with equal heights are proportional to the bases of the triangles. Referring to Figure 3, we have:
|BX||XC|= SABXSAXC= SPBXSPXC= SABX−SPBXSAXC−SPXC= SABPSCAP.
Likewise,
|CY||YA|= SBCPSABP, |AZ||ZB|= SCAPSBCP.
Now if we multiply them we get
|BX||XC|· |CY||YA|· |AZ||ZB|= SABPSCAP· SBCPSABP· SCAPSBCP=1 .
The converse of this theorem is also true:
Theorem 1.22. If three cevians AX, BY, CZ satisfy the relation
|BX||XC|· |CY||YA|· |AZ||ZB|=1 ,
then they are competitive .
To show this, suppose that the first two cevians intersect at the pointP , as before, and the third cevian passing through the pointP , willCZ′ . Then, by Theorem 1.21,
|BX||XC|· |CY||YA|· |AZ′||Z′B|=1 .
But by assumption
|BX||XC|· |CY||YA|· |AZ||ZB|=1 .
Hence,
|AZ||ZB|= |AZ′||Z′B| ,
dotZ′ coincides with the pointZ , and we proved that the segmentsAX , BY AndCZ competitive (, p. 54 and , pp. 48, 317).
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.
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 thing 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.
— What do Menelaus’ theorem and drugs have in common?
“Everyone knows about them, but no one talks about them.”
Typical conversation with a student
This is a cool theorem that will help you at a time when it seems that nothing can help. In this lesson we will formulate the theorem itself, consider several options for its use, and as a dessert you will have some tough homework. Go!
First, the wording. Perhaps I will not give the most “beautiful” version of the theorem, but the most understandable and convenient.
Menelaus's theorem. Let's consider an arbitrary triangle $ABC$ and a certain line $l$ that intersects two sides of our triangle internally and one side on the continuation. Let us denote the intersection points of $M$, $N$ and $K$:
Triangle $ABC$ and secant $l$
Then the following relation is true:
\[\frac(AM)(MB)\cdot \frac(BN)(NC)\cdot \frac(CK)(KA)=1\]
I would like to note: there is no need to cram the placement of letters in this evil formula! Now I will tell you an algorithm by which you can always restore all three fractions literally on the fly. Even during an exam under stress. Even if you are sitting at geometry at 3 am and don’t understand anything at all. :)
The scheme is simple:
- Draw a triangle and a secant. For example, as shown in the theorem. We designate vertices and points with some letters. It can be an arbitrary triangle $ABC$ and a straight line with points $M$, $N$, $K$, or some other one - that’s not the point.
- Place a pen (pencil, marker, quill pen) at any vertex of the triangle and begin traversing the sides of this triangle with obligatory entry into the points of intersection with the straight line. For example, if we first go from point $A$ to point $B$, we will get the segments: $AM$ and $MB$, then $BN$ and $NC$, and then (attention!) $CK$ and $KA$ . Since point $K$ lies on the continuation of side $AC$, when moving from $C$ to $A$ you will have to temporarily leave the triangle.
- And now we simply divide the adjacent segments into each other exactly in the order in which we received them when traversing: $AM/MB$, $BN/NC$, $CK/KA$ - we get three fractions, the product of which will give us one .
In the drawing it will look like this:
A simple scheme that allows you to restore the formula from Menelaus And just a couple of comments. More precisely, these are not even comments, but answers to typical questions:
- What happens if line $l$ passes through the vertex of the triangle? Answer: nothing. Menelaus' theorem does not work in this case.
- What happens if you choose another vertex to start or go in the other direction? Answer: it will be the same. The sequence of fractions will simply change.
I think we've sorted out the wording. Let's see how all this stuff is used to solve complex geometric problems.
Why is all this needed?
Warning. Excessive use of Menelaus' theorem to solve planimetric problems can cause irreparable harm to your psyche, since this theorem significantly speeds up calculations and forces you to remember other important facts from a school geometry course.
Proof
I won't prove it. :)
Okay, I'll prove it:
Now it remains to compare the two obtained values for the segment $CT$:
\[\frac(AM\cdot BN\cdot CK)(BM\cdot CN\cdot AK)=1;\]
\[\frac(AM)(BM)\cdot \frac(BN)(CN)\cdot \frac(CK)(AK)=1;\]
OK it's all over Now. All that remains is to “comb” this formula by correctly placing the letters inside the segments - and the formula is ready. :)
