Lesson "theorems on angles formed by two parallel lines and a transversal." Material on mathematics "theorems on angles formed by chords, tangents and secants"

A video lesson about theorems on angles between two parallel lines and their transversal contains material presenting the structural features of the theorem, examples of the formation and proof of converse theorems, and corollaries from them. The purpose of this video lesson is to deepen the concept of a theorem, decomposing it into its components, considering the concept of an inverse theorem, to develop the ability to construct a theorem inverse to a given theorem, consequences from the theorem, and to develop the ability to prove statements.

The form of the video lesson allows you to successfully place emphasis when demonstrating the material, making it easier to understand and remember the material. The topic of this video lesson is complex and important, so the use of a visual aid is not only advisable, but also desirable. It provides an opportunity to improve the quality of learning. Animated effects enhance the presentation educational material, bring the learning process closer to the traditional one, and the use of video frees the teacher to deepen individual work.

The video lesson begins with the announcement of its topic. At the beginning of the lesson, the decomposition of the theorem into its components is considered for a better understanding of its structure and possibilities for further research. A diagram is shown on the screen demonstrating that the theorem consists of its conditions and conclusions. The concept of condition and conclusion is described using the example of the sign of parallel lines, noting that part of the statement is the condition of the theorem, and the conclusion is the conclusion.

Deepening the knowledge gained about the structure of the theorem, students are given the concept of a theorem inverse to a given one. It is formed as a result of replacement - the condition becomes the conclusion, the conclusion - the condition. To develop students’ ability to construct theorems inverse to data and the ability to prove them, theorems are considered, reverse topics, which are discussed in lesson 25 about the signs of parallel lines.

The screen displays the theorem inverse to the first theorem, which describes the sign of parallel lines. By interchanging the condition and conclusion, we obtain the statement that if any parallel lines are intersected by a transversal, then the crosswise angles formed in this case will be equal. The proof is demonstrated in the figure, which shows lines a, b, as well as a transversal passing through these lines at their points M and N. The crosswise angles ∠1 and ∠2 are marked in the image. It is necessary to prove their equality. First, the proof makes the assumption that these angles are not equal. To do this, a certain straight line P is drawn through the point M. An angle `∠PMN is constructed, which lies crosswise with an angle ∠2 with respect to MN. The angles `∠PMN and ∠2 are equal by construction, therefore MP║b. Conclusion - two lines parallel to a point are drawn through b. However, this is impossible because it does not correspond to the parallel lines axiom. The assumption made turns out to be wrong, proving the validity of the original statement. The theorem has been proven.

Next, students' attention is drawn to the method of proof that was used in the course of the reasoning. A proof in which the assertion being proved is assumed to be false is called proof by contradiction in geometry. This method is often used to prove various geometric statements. IN in this case, having assumed the inequality of cross-lying angles, a contradiction emerged in the course of the reasoning, which denies the validity of such a contradiction.

Students are reminded that a similar method has been used before in proofs. An example of this is the proof of the theorem in lesson 12 that two lines that are perpendicular to a third do not intersect, as well as the proof of the corollaries in lesson 28 from the axiom of parallel lines.

Another provable corollary states that a line is perpendicular to both parallel lines if it is perpendicular to one of them. The figure shows straight lines a and b and a straight line c perpendicular to them. The perpendicularity of the straight line c to a means that the angle formed with it is equal to 90°. The parallelism of a and b and their intersection with line c means that line c intersects b. The angle ∠2 formed with the line b is crosswise to the angle ∠1. And since, according to the condition, the lines are parallel, then these angles are equal. Accordingly, the angle ∠2 will also be equal to 90°. This means that line c is perpendicular to line b. The theorem under consideration has been proven.

Next we prove the theorem converse to the second criterion for parallel lines. The converse theorem states that if two straight lines are parallel, the corresponding angles formed will be equal. The proof begins with the construction of a secant c and parallel lines a and b. The angles created in this case are marked in the figure. There are a pair of corresponding angles called ∠1 and ∠2, and also marked angle ∠3, which lies crosswise with angle ∠1. The parallelism of a and b means the equality ∠3=∠1 as lying crosswise. Considering that ∠3, ∠2 are vertical, they are also equal. A consequence of such equalities is the statement that ∠1=∠2. The theorem under consideration has been proven.

The last one provable on this lesson the theorem is the inverse of the last criterion for parallel lines. Its text states that if a transversal passes through parallel lines, the sum of the one-sided angles formed is equal to 180°. The progress of the proof is demonstrated in the figure, which shows lines a and b intersecting the secant c. It is necessary to prove that the sum of one-sided angles will be equal to 180°, that is, ∠4+∠1 = 180°. From the parallelism of straight lines a and b, the equality of the corresponding angles ∠1 and ∠2 follows. The adjacency of angles ∠4, ∠2 means that they add up to 180°. In this case, the angles ∠1= ∠2 - this means that ∠1 added to the angle ∠4 will be 180°. The theorem has been proven.

For a deeper understanding of how inverse theorems are formed and proven, it is separately noted that if a theorem is proven and true, this does not mean that the inverse theorem will also be true. To understand this, a simple example is given. There is a theorem that all vertical angles are equal. The converse theorem sounds like all equal angles are vertical, which is not true. After all, you can construct two equal angles that are not vertical. This can be seen in the picture shown.

The video lesson “Theorems on angles formed by two parallel lines and a transversal” is visual aid, which can be used by a teacher in a geometry lesson, and can also successfully form an idea of ​​inverse theorems and corollaries, as well as their proof in self-study material, to be useful in distance learning.

Theorem: If two parallel lines are intersected by a transversal, then the intersecting angles are equal. and in A B = 2 s

Proof: A B CD M N 1 2 A B CD M N 1 2 K O Let the lines AB and CD be parallel, MN their secant. Let us prove that the crosswise angles 1 and 2 are equal to each other. Let's assume that 1 and 2 are not equal. Let us draw a straight line KF through point O. Then at point O it is possible to construct KON, lying crosswise and equal to 2. But if KON = 2, then the straight line KF will be parallel to CD. We found that two straight lines AB and KF are drawn through point O, parallel to straight line CD. But this cannot be. We arrived at a contradiction because we assumed that 1 and 2 are not equal. Therefore, our assumption is incorrect and 1 must be equal to 2, i.e., crosswise angles are equal. F

Theorem: If two parallel lines are intersected by a transversal, then the corresponding angles are equal. and in A B = 2

Theorem: If two parallel lines are intersected by a transversal, then the sum of the one-sided angles is 180°. and in A B = 180°

Proof: Let parallel lines a and b be intersected by secant AB, then the corresponding 1 and 2 will be equal, 2 and 3 will be adjacent, therefore = 180°. From the equalities 1 = 2 and = 180° it follows that = 180°. The theorem has been proven. 2 a in A B 3 1

Solution: 1. Let X be 2, then 1 = (X+70°), because the sum of angles 1 and 2 = 180°, due to the fact that they are adjacent. Let's make an equation: X+ (X+70°) = 180° 2X = 110° X = 55° (Angle 2) 2. Find 1. 55° + 70° = 125° 3. 1 = 3, because they are vertical. 3 = 5, because they are lying crosswise. 125° 5 = 7, because they are vertical. 2 = 4, because they are vertical. 4 = 6, because they are lying crosswise. 55° 6 = 8, because they are vertical. Problem 1: A B Condition: Find all the angles formed when two parallel lines A and B intersect with a transversal C, if one of the angles is 70° greater than the other.

Solution: 1. 1= 2, because they are vertical, which means 2= 45° is adjacent to 2, so 3+ 2=180°, and it follows from this that 3= 180° - 45°= 135° = 180°, because they are one-sided. 4 = 45°. Answer: 4=45°; 3=135°. Problem 3: A B 2 Condition: two parallel lines A and B are intersected by a secant C. Find what 4 and 3 will be equal to if 1=45°

Theorems about angles formed

Geometry, Chapter III, 7th grade

To the textbook by L.S. Atanasyan

mathematics teacher of the highest category

Municipal educational institution "Upshinskaya basic secondary school"

Orsha district of the Republic of Mari El

The converse of this theorem

Theorem: In an isosceles triangle, the base angles are equal .

Theorem: If a triangle is isosceles, then its base angles are equal .

Condition of the theorem (Given): triangle - isosceles

Conclusion of the theorem (Prove): base angles are equal

Condition of the theorem : base angles are equal

Conclusion of the theorem : triangle - isosceles

NEW STATEMENT

Reverse

theorem

If a triangle has two angles

are equal, then it is isosceles .

The converse of this theorem

Is the converse always true?

Theorem

Converse theorem

If the sum of two angles is 180 0 , then the angles are adjacent

Sum of adjacent angles

equal to 180 0 .

If the angles are equal,

then they are vertical

Vertical angles are equal

If in a triangle the bisector drawn to one of its sides is also the median drawn to this side, then this triangle is isosceles

In an isosceles triangle, the bisector drawn to the base is the median and altitude

If in a triangle the bisector drawn to one of its sides is also the altitude drawn to this side, then this triangle is isosceles

E If the triangle is isosceles, then the bisector drawn to the base , is both the median and the height

Angles formed by two parallel lines and a transversal

Is the converse always true?

Theorem

Converse theorem

If two parallel lines are crossed by a secant, then crossed angles are equal

crosswise angles equal That lines are parallel .

But this contradicts axiom of parallel , then our assumption is incorrect

FROM METHOD

OPPOSITE

We put forward an assumption opposite to what needs to be proven

Through reasoning we come to a contradiction with a well-known axiom or theorem

We conclude that our assumption is incorrect and the theorem is correct

But this contradicts axiom of parallel

Therefore, our assumption is incorrect

If two parallel lines are intersected by a transversal, then the intersecting angles are equal

COROLLARY FROM THE THEOREM

If a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other

Angles formed

two parallel lines and a transversal

Theorem

Converse theorem

If, at the intersection of two straight lines, a secant corresponding angles are equal , That lines are parallel .

If two parallel lines are crossed by a secant, then corresponding angles are equal

Angles formed

two parallel lines and a transversal

Theorem

Converse theorem

If, at the intersection of two straight lines, a secant 0 , That lines are parallel .

If two parallel lines are crossed by a secant, then the sum of one-sided angles is 180 0

Lines a and b are parallel.

Find angle 2.

Lines a and b are parallel.

Find unknown angles

Lines a and b are parallel.

Find unknown angles

Find unknown angles

Find unknown angles

Find unknown angles

Lines a and b are parallel. Find the unknown angles if the sum of two intersecting angles is 100 0 .

Lines a and b are parallel. Find the unknown angles if the sum of two corresponding angles is 260 0 .

Lines a and b are parallel. Find the unknown angles if the difference between two one-sided angles is 50 0 .

\[(\Large(\text(Central and inscribed angles)))\]

Definitions

A central angle is an angle whose vertex lies at the center of the circle.

An inscribed angle is an angle whose vertex lies on a circle.

The degree measure of an arc of a circle is the degree measure of the central angle that subtends it.

Theorem

The degree measure of an inscribed angle is equal to half the degree measure of the arc on which it rests.

Proof

We will carry out the proof in two stages: first, we will prove the validity of the statement for the case when one of the sides of the inscribed angle contains a diameter. Let point \(B\) be the vertex of the inscribed angle \(ABC\) and \(BC\) be the diameter of the circle:

Triangle \(AOB\) is isosceles, \(AO = OB\) , \(\angle AOC\) is external, then \(\angle AOC = \angle OAB + \angle ABO = 2\angle ABC\), where \(\angle ABC = 0.5\cdot\angle AOC = 0.5\cdot\buildrel\smile\over(AC)\).

Now consider an arbitrary inscribed angle \(ABC\) . Let us draw the diameter of the circle \(BD\) from the vertex of the inscribed angle. There are two possible cases:

1) the diameter cuts the angle into two angles \(\angle ABD, \angle CBD\) (for each of which the theorem is true as proven above, therefore it is also true for the original angle, which is the sum of these two and therefore equal to half the sum of the arcs to which they rest, that is, equal to half the arc on which it rests). Rice. 1.

2) the diameter did not cut the angle into two angles, then we have two more new inscribed angles \(\angle ABD, \angle CBD\), whose side contains the diameter, therefore, the theorem is true for them, then it is also true for the original angle (which is equal to the difference of these two angles, which means it is equal to the half-difference of the arcs on which they rest, that is, equal to half the arc on which it rests). Rice. 2.

Consequences

1. Inscribed angles subtending the same arc are equal.

2. An inscribed angle subtended by a semicircle is a right angle.

3. An inscribed angle is equal to half the central angle subtended by the same arc.

\[(\Large(\text(Tangent to the circle)))\]

Definitions

There are three types of relative positions of a line and a circle:

1) straight line \(a\) intersects the circle at two points. Such a line is called a secant. In this case, the distance \(d\) from the center of the circle to the straight line is less than the radius \(R\) of the circle (Fig. 3).

2) straight line \(b\) intersects the circle at one point. Such a line is called a tangent, and their common point \(B\) is called the point of tangency. In this case \(d=R\) (Fig. 4).

Theorem

1. A tangent to a circle is perpendicular to the radius drawn to the point of tangency.

2. If a line passes through the end of the radius of a circle and is perpendicular to this radius, then it is tangent to the circle.

Consequence

The tangent segments drawn from one point to a circle are equal.

Proof

Let us draw two tangents \(KA\) and \(KB\) to the circle from the point \(K\):

This means that \(OA\perp KA, OB\perp KB\) are like radii. Right Triangles\(\triangle KAO\) and \(\triangle KBO\) are equal in leg and hypotenuse, therefore, \(KA=KB\) .

Consequence

The center of the circle \(O\) lies on the bisector of the angle \(AKB\) formed by two tangents drawn from the same point \(K\) .

\[(\Large(\text(Theorems related to angles)))\]

Theorem on the angle between secants

The angle between two secants drawn from the same point is equal to the half-difference in degree measures of the larger and smaller arcs they cut.

Proof

Let \(M\) be the point from which two secants are drawn as shown in the figure:

Let's show that \(\angle DMB = \dfrac(1)(2)(\buildrel\smile\over(BD) - \buildrel\smile\over(CA))\).

\(\angle DAB\) is the external angle of the triangle \(MAD\), then \(\angle DAB = \angle DMB + \angle MDA\), where \(\angle DMB = \angle DAB - \angle MDA\), but the angles \(\angle DAB\) and \(\angle MDA\) are inscribed, then \(\angle DMB = \angle DAB - \angle MDA = \frac(1)(2)\buildrel\smile\over(BD) - \frac(1)(2)\buildrel\smile\over(CA) = \frac(1)(2)(\buildrel\smile\over(BD) - \buildrel\smile\over(CA))\), which was what needed to be proven.

Theorem on the angle between intersecting chords

The angle between two intersecting chords is equal to half the sum of the degree measures of the arcs they cut: \[\angle CMD=\dfrac12\left(\buildrel\smile\over(AB)+\buildrel\smile\over(CD)\right)\]

Proof

\(\angle BMA = \angle CMD\) as vertical.

From triangle \(AMD\) : \(\angle AMD = 180^\circ - \angle BDA - \angle CAD = 180^\circ - \frac12\buildrel\smile\over(AB) - \frac12\buildrel\smile\over(CD)\).

But \(\angle AMD = 180^\circ - \angle CMD\), from which we conclude that \[\angle CMD = \frac12\cdot\buildrel\smile\over(AB) + \frac12\cdot\buildrel\smile\over(CD) = \frac12(\buildrel\smile\over(AB) + \buildrel\ smile\over(CD)).\]

Theorem on the angle between a chord and a tangent

The angle between the tangent and the chord passing through the point of tangency is equal to half the degree measure of the arc subtended by the chord.

Proof

Let the straight line \(a\) touch the circle at the point \(A\), \(AB\) is the chord of this circle, \(O\) is its center. Let the line containing \(OB\) intersect \(a\) at the point \(M\) . Let's prove that \(\angle BAM = \frac12\cdot \buildrel\smile\over(AB)\).

Let's denote \(\angle OAB = \alpha\) . Since \(OA\) and \(OB\) are radii, then \(OA = OB\) and \(\angle OBA = \angle OAB = \alpha\). Thus, \(\buildrel\smile\over(AB) = \angle AOB = 180^\circ - 2\alpha = 2(90^\circ - \alpha)\).

Since \(OA\) is the radius drawn to the tangent point, then \(OA\perp a\), that is, \(\angle OAM = 90^\circ\), therefore, \(\angle BAM = 90^\circ - \angle OAB = 90^\circ - \alpha = \frac12\cdot\buildrel\smile\over(AB)\).

Theorem on arcs subtended by equal chords

Equal chords subtend equal arcs smaller than semicircles.

And vice versa: equal arcs are subtended by equal chords.

Proof

1) Let \(AB=CD\) . Let us prove that the smaller semicircles of the arc .

On three sides, therefore, \(\angle AOB=\angle COD\) . But because \(\angle AOB, \angle COD\) - central angles, resting on arcs \(\buildrel\smile\over(AB), \buildrel\smile\over(CD)\) accordingly, then \(\buildrel\smile\over(AB)=\buildrel\smile\over(CD)\).

2) If \(\buildrel\smile\over(AB)=\buildrel\smile\over(CD)\), That \(\triangle AOB=\triangle COD\) on two sides \(AO=BO=CO=DO\) and the angle between them \(\angle AOB=\angle COD\) . Therefore, and \(AB=CD\) .

Theorem

If the radius bisects the chord, then it is perpendicular to it.

The converse is also true: if the radius is perpendicular to the chord, then at the point of intersection it bisects it.

Proof

1) Let \(AN=NB\) . Let us prove that \(OQ\perp AB\) .

Consider \(\triangle AOB\) : it is isosceles, because \(OA=OB\) – radii of the circle. Because \(ON\) is the median drawn to the base, then it is also the height, therefore, \(ON\perp AB\) .

2) Let \(OQ\perp AB\) . Let us prove that \(AN=NB\) .

Similarly, \(\triangle AOB\) is isosceles, \(ON\) is the height, therefore, \(ON\) is the median. Therefore, \(AN=NB\) .

\[(\Large(\text(Theorems related to the lengths of segments)))\]

Theorem on the product of chord segments

If two chords of a circle intersect, then the product of the segments of one chord is equal to the product of the segments of the other chord.

Proof

Let the chords \(AB\) and \(CD\) intersect at the point \(E\) .

Consider the triangles \(ADE\) and \(CBE\) . In these triangles, angles \(1\) and \(2\) are equal, since they are inscribed and rest on the same arc \(BD\), and angles \(3\) and \(4\) are equal as vertical. Triangles \(ADE\) and \(CBE\) are similar (based on the first criterion of similarity of triangles).

Then \(\dfrac(AE)(EC) = \dfrac(DE)(BE)\), from which \(AE\cdot BE = CE\cdot DE\) .

Tangent and secant theorem

The square of a tangent segment is equal to the product of a secant and its outer part.

Proof

Let the tangent pass through the point \(M\) and touch the circle at the point \(A\) . Let the secant pass through the point \(M\) and intersect the circle at the points \(B\) and \(C\) so that \(MB< MC\) . Покажем, что \(MB\cdot MC = MA^2\) .

Consider the triangles \(MBA\) and \(MCA\) : \(\angle M\) is common, \(\angle BCA = 0.5\cdot\buildrel\smile\over(AB)\). According to the theorem about the angle between a tangent and a secant, \(\angle BAM = 0.5\cdot\buildrel\smile\over(AB) = \angle BCA\). Thus, triangles \(MBA\) and \(MCA\) are similar at two angles.

From the similarity of triangles \(MBA\) and \(MCA\) we have: \(\dfrac(MB)(MA) = \dfrac(MA)(MC)\), which is equivalent to \(MB\cdot MC = MA^2\) .

Consequence

The product of a secant drawn from the point \(O\) by its external part does not depend on the choice of the secant drawn from the point \(O\) .