Teaching planimetry in a school course

The theorem on the property of angles with correspondingly parallel sides should be considered for cases where the given angles are either both acute, or both obtuse, or one of them is acute and the other is obtuse.

The theorem finds wide application when studying the properties of various figures and, in particular, the quadrilateral.

The indication that the sides of angles with correspondingly parallel sides can have either the same or opposite direction, which is sometimes found in the formulation of theorems, is considered unnecessary. If we use the term “direction,” then it would be necessary to clarify what should be understood by this word. It is enough to draw students' attention to the fact that angles with correspondingly parallel sides are equal if they are both acute or both obtuse, but if one of the angles is obtuse and the other acute, then they add up to 2d.

The theorem on angles with correspondingly perpendicular sides can be given immediately after the theorem on the property of angles with correspondingly parallel sides. Students are given examples of using the properties of angles with parallel and perpendicular sides, respectively, in devices and machine parts.

Sum of triangle angles

When deriving the theorem on the sum of the angles of a triangle, you can use visual aids. Triangle ABC is cut out, its corners are numbered, then they are cut off and applied to each other. It turns out l+2+3=2d. Draw height CD from vertex C of triangle ABC and bend the triangle so that the height is divided in half, i.e. vertex C fell to point D - the base of the height. The inflection line MN is the midline of triangle ABC. Then they bend isosceles triangles AMD and DNB by their heights, with vertices A and B coinciding with point D and l+2+3=2d.

It should be remembered that the use of visual aids in a systematic course of geometry is not intended to replace the logical proof of a proposition with its experimental verification. Visual aids should only facilitate students’ understanding of this or that geometric fact, the properties of this or that geometric figure and the relative position of its individual elements. When determining the size of the angle of a triangle, students should be reminded of the previously discussed theorem on the external angle of a triangle and indicate that the theorem on the sum of the angles of a triangle allows, both by construction and calculation, to establish a numerical relationship between external and internal angles that are not adjacent to them.

As a consequence of the theorem on the sum of the angles of a triangle, it is proved that in a right triangle the leg opposite the angle of 30 degrees is equal to half the hypotenuse.

As the material progresses, students should ask questions and simple tasks, promoting better assimilation of new material. For example, Which lines are called parallel?

At what position of the transversal are all angles formed by two parallel lines and this transversal equal?

A straight line drawn in a triangle parallel to the base cuts off a small triangle from it. Prove that the triangle being cut off and the given triangle are congruent.

Calculate all the angles subtended by two parallels and a transversal if it is known that one of the angles is 72 degrees.

The internal one-sided angles are respectively equal to 540 and 1230. By how many degrees should one of the lines be rotated around the point of its intersection with the transversal so that the lines are parallel?

Prove that the bisectors of: a) two equal but not opposite angles formed by two parallel lines and a transversal are parallel, b) two unequal angles with the same lines and a transversal are perpendicular.

Given two parallel lines AB and CD and a secant EF intersecting these lines at points K and L. The drawn bisectors KM and KN of the angles AKL and BKL cut off the segment MN on the straight line CD. Find the length MN if it is known that the secant segment KL enclosed between the parallel ones is equal to a.

What is the type of triangle in which: a) the sum of any two angles is greater than d, b) the sum of two angles is equal to d, c) the sum of two angles is less than d? Answer: a) acute-angled, b) rectangular, c) obtuse-angled. How many times is the sum of the exterior angles of a triangle more than the amount its internal corners? Answer: 2 times.

Can all the external angles of a triangle be: a) acute, b) obtuse, c) straight? Answer: a) no, b) yes, c) no.

Which triangle has each exterior angle twice the size of each interior angle? Answer: equilateral.

When studying the technique of parallel lines, it is necessary to use historical, theoretical and methodological literature to fully formulate the concept of parallel lines.

THEOREM 1.Equality of angles with mutually perpendicular sides:If
both sharp or both obtuse and
,
, That
. THEOREM 2. Properties of the midline of a trapezoid:A) the midline of the trapezoid is parallel to the bases of the trapezoid;B) the middle line is equal to half the sum of the bases of the trapezoid;C) the middle line (and only it) bisects any segment enclosed between the bases of the trapezoid. These properties are also valid for the midline of a triangle, if we consider the triangle to be a “degenerate” trapezium, one of the bases of which has a length equal to zero. THEOREM 3. On the intersection points of medians, bisectors, altitudes of a triangle:A) three medians of a triangle intersect at one point (it is called the center of gravity of the triangle) and divide at this point in a ratio of 2: 1, counting from the vertex;B) three bisectors of a triangle intersect at one point;C) three altitudes intersect at one point (it is called the orthocenter of the triangle).THEOREM 4. Property of the median in a right triangle:in a right triangle, the median drawn to the hypotenuse is equal to half of it. The converse theorem is also true: if in a triangle one of the medians is equal to half the side to which it is drawn, then this triangle is right angledTHEOREM 5. property of the bisector of an internal angle of a triangle:The bisector of an internal angle of a triangle divides the side to which it is drawn into parts proportional to the opposite sides:
THEOREM 6. Metric relations in a right triangle:IfaAndb– legs,c– hypotenuse,h- height, And - projections of the legs onto the hypotenuse, then: a)
; b)
; V)
; G)
; d)
THEOREM 7. Determination of the type of triangle based on its sides:Leta, b, c– sides of the triangle, with c being the largest side; Then:And if
, then the triangle is acute;B) if
, then the triangle is right-angled;B) if
, then the triangle is obtuse.THEOREM 8. Metric relations in a parallelogram:The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of all its sides:
. When solving geometric problems, you often have to establish the equality of two segments (or angles). Let's indicate three main ways of geometrically proving the equality of two segments: 1) consider the segments as sides of two triangles and prove that these triangles are equal; 2) represent the segments as sides of a triangle and prove that this triangle is isosceles; 3 ) replace the segment A an equal segment , and the segment b – equal to it and prove the equality of the segments and . Task 1.Two mutually perpendicular lines intersect the sidesAB, B.C., CD, ADsquareABCDat pointsE, F, K, Lrespectively. Prove thatE.K. = FL(see figure for task No. 1).R

Rice. to task No. 1

Solution: 1. Using the first of the above paths for the equality of two segments, we draw the segments
And
- then the segments we are interested in E.K. And FL become sides of two right triangles EPK And FML(see figure for task No. 1) . 2

Rice. to task No. 1

We have: PK = FM(more details: PK = AD, AD = AB, AB = FM, Means,PK = FM), (as angles with mutually perpendicular sides, Theorem 1). This means (along the leg and acute angle). From the equality of right triangles it follows that their hypotenuses are equal, i.e. segments E.K. And FL. ■ Note that when solving geometric problems, you often have to make additional constructions, for example the following: drawing a straight line parallel or perpendicular to one of those in the figure (as we did in task 1); doubling the median of the triangle in order to complete the triangle to a parallelogram (we will do this in Problem 2), drawing an auxiliary bisector. There are useful additional constructions related to the circle. Task 2.Parties
equala, b, c. Calculate median , drawn to side c. (see figure for problem 2).R

Rice. to problem No. 2

Solution: Double the median by completing
to the parallelogram ACVR, and apply Theorem 8 to this parallelogram. We obtain: , i.e.
, where we find:
Task 3.Prove that in any triangle the sum of the medians is greater than ¾ of the perimeter, but less than the perimeter.R
solution: 1. Let's consider
(see figure for problem 3) We have:
;
. Because AM + MS > AC, That
(1) P

Rice. to problem No. 3

Carrying out similar reasoning for triangles AMB and BMC, we obtain:
(2)
(3) Adding inequalities (1), (2), (3), we obtain:
, T
.e. we proved that the sum of the medians is greater than ¾ of the perimeter. 2. Let's double the median BD, completing the triangle to a parallelogram (see figure for problem 3). Then from
we get: B.K. < B.C. + CK, those.
(4) Likewise:
(5)

Rice. to problem No. 3

(6) Adding inequalities (4), (5), (6), we obtain: , i.e. the sum of the medians is less than the perimeter. ■ Task 4.Prove that in a non-isosceles right triangle the bisector right angle bisects the angle between the median and the altitude drawn from the same vertex.R
solution: Let ACB be a right triangle,
, CH – height, CD – bisector, SM – median. Let us introduce the following notation: (see figure for problem 4) . 1.
as angles with mutually perpendicular sides (). 2

Rice. to problem No. 4

Because
(see Theorem 4), then SM = MV, and then from
we conclude that
So, 3. Since and (after all, CD is a bisector), that is what needed to be proved. ■ Task 5.In a parallelogram with sidesa Andbbisectors of internal angles are drawn (see figure for problem 5). Find the lengths of the diagonals of the quadrilateral formed at the intersection of the bisectors.Solution: 1 . AE – bisector
, BP – bisector
(see figure) . since in a parallelogram
those. then This means that in triangle ABC the sum of angles A and B is equal to 90 0, then angle K is equal to 90 0, i.e., the bisectors AE and BP are mutually perpendicular. A
The mutual perpendicularity of the bisectors AE and DQ, BP and CF, CF and DQ is proved logically. OUTPUT: KLMN is a quadrilateral with right angles, i.e. rectangle. A rectangle has equal diagonals, so it is enough to find the length of one of them, for example KM. 2

Rice. to problem No. 5

Let's consider
He has AK - both bisector and height. This means, firstly, the triangle ABP is isosceles, i.e. AB = AP = b, and, secondly, that the segment AK is at the same time the median of the triangle ABP, i.e. K – the middle of the bisector BP. It is proved in a similar way that M is the midpoint of the bisector DQ. 3. Let's consider the segment KM. It bisects the segments BP and DQ. But the middle line of a parallelogram (note that a parallelogram is special case trapezoid; If we can talk about the midline of a trapezoid, then we can equally well talk about the midline of a parallelogram, which has the same properties) passes through points K and M (see Theorem 2). This means that KM is a segment on the midline, and therefore
.4. Because
And
, then KMDP is a parallelogram, and therefore. Answer:
■ In fact, in the process of solving the problem (at stages 1 and 2) we proved quite important property: the bisectors of angles adjacent to the side of a trapezoid intersect at right angles at a point lying on the midline of the trapezoid. It should be noted that the main method of composing equations in geometric problems is methodsupport element, which is as follows: the same element (side, angle, area, radius, etc.) is expressed through known and unknown quantities by two different ways and the resulting expressions are equated. Quite often, an area is chosen as a reference elementfigures. Then we say that to construct the equation we use area method. It is necessary to teach schoolchildren how to solve basic problems, i.e. those. Which are included as components in many other tasks. These are, for example, problems of finding the basic elements of a triangle: median, height, bisector, inscribed and circumscribed circle radii, area. Z problem 6.In triangle ABC, sides AB and BC are equal, and BH is the height. A point is taken on the BC sideDSo
(see figure for problem 6). In what ratio is the segmentADdivides the height of the VN?Solution: 1. Let BD = a, then CD = 4 a, AB = 5a.2

Rice. to problem No. 6

Let's draw a segment
(see figure for problem 6) Since NK is the middle line of the triangle ACD DK = KC = 2 a .3. Consider the triangle VNK. We have: BD = a,DK = 2 a And
. According to Thales' theorem
But
That means
■ If the problem requires finding the ratio of any number of quantities, then, as a rule, the problem is solved using the auxiliary parameter method. This means that at the beginning of solving a problem we declare some linear value known, denoting it, for example, with the letter A, and then express it through A those quantities whose ratio is required to be found. When the required relation is constructed, the auxiliary parameter A is shrinking. This is exactly how we acted in the problem . Our advice: when solving problems in which it is necessary to find the ratio of quantities (in particular, in problems of determining an angle - after all, as a rule, when calculating an angle we're talking about about finding him trigonometric function, i.e. about the relationship of the parties right triangle), students should be taught to highlight the introduction of an auxiliary parameter as the first stage of solving. The auxiliary parameter method is also used in problems where geometric figure determined up to similarity. Task 7.A rectangle is inscribed in a triangle with sides equal to 10, 17 and 21 cm so that its two vertices are on one side of the triangle, and the other two vertices are on the other two sides of the triangle. Find the sides of the rectangle if it is known that its perimeter is 22.5 cm.R
decision . 1. First of all, let's determine the type of triangle. We have: 10 2 = 100; 17 2 = 289; 21 2 = 441. Since 21 2 > 10 2 + 17 2, the triangle is obtuse-angled (see Theorem 7), which means that a rectangle can be inscribed in it only in one way: by placing its two vertices on the larger side of triangle ABC (see fig. . to problem 7), where AC = 21 cm, AB = 10 cm, BC = 17 cm. 2

Typically, angles are considered either with corresponding parallel sides or with corresponding perpendicular sides. Let's consider the first case first.

Let two angles ABC and DEF be given. Their sides are respectively parallel: AB || DE and BC || E.F. These two angles will either be equal or their sum will be equal to 180°. In the figure below, in the first case ∠ABC = ∠DEF, and in the second ∠ABC + ∠DEF = 180°.

The proof that this is indeed the case comes down to the following.

Consider angles with correspondingly parallel sides, located as in the first figure. At the same time, we extend straight lines AB and EF until they intersect. Let us denote the intersection point by the letter G. In addition, for clarity of the subsequent proof, side BC is extended in the figure.

Since lines BC and EF are parallel, if line AB intersects one of them, then it will certainly intersect the other. That is, line AB is a secant for two parallel lines. As is known, in this case, the crosswise angles at the secant are equal, one-sided angles add up to 180°, and the corresponding angles are equal.

That is, no matter what pair of angles we take at vertices B and G (one angle from one, the other from the second), we will always get either equal angles or adding up to 180°.

However, lines AB and DE are also parallel. For them, the straight line EF is a secant. This means that any pairs of angles from vertices G and E will add up to either 180° or be equal to each other. It follows that pairs of angles from vertices B and E will obey this rule.

For example, consider the angles ∠ABC and ∠DEF. Angle ABC equal to angle BGE, since these angles are corresponding to parallel lines BC and EF. In turn, angle BGE is equal to angle DEF, since these angles are corresponding when AB and DE are parallel. Thus it is proved, ∠ABC and ∠DEF.

Now consider the angles ∠ABC and ∠DEG. Angle ABC is equal to angle BGE. But ∠BGE and ∠DEG are one-sided angles with parallel lines (AB || DE) intersected by a transversal (EF). As you know, such angles add up to 180°. If we look at the second case in the first picture, we realize that it corresponds to the pair of angles ABC and DEG in the second picture.

So two different angles, whose sides are respectively parallel, or equal to each other, or add up to 180°. The theorem has been proven.

A special case should be noted - when the corners are turned. In this case, they will obviously be equal to each other.

Now consider angles with correspondingly perpendicular sides. This case looks more complicated because mutual arrangement angles are more varied. The figure below shows three examples of how corners can be positioned with correspondingly perpendicular sides. However, in either case, one side of the first angle (or its extension) is perpendicular to one side of the second angle, and the second side of the first angle is perpendicular to the second side of the second angle.

Let's consider one of the cases. In this case, we draw a bisector in one corner and through an arbitrary point of it we draw perpendiculars to the sides of its angle.

Here are angles ABC and DEF with respectively perpendicular sides: AB ⊥ DE and BC ⊥ EF. On the bisector of angle ABC, point G is taken, through which perpendiculars to the same angle are drawn: GH ⊥ AB and GI ⊥ BC.

Consider triangles BGH and BGI. They are rectangular because the angles H and I are right angles. In them, the angles at vertex B are equal, since BG is the bisector of angle ABC. Also, the side BG of the triangles under consideration is common and is the hypotenuse for each of them. As is known, right triangles are congruent if their hypotenuses and one of them are equal. sharp corners. Thus, ∆BGH = ∆BGI.

Since ∆BGH = ∆BGI, then ∠BGH = ∠BGI. Therefore, angle HGI can be represented not as the sum of these two angles, but as one of them multiplied by 2: ∠HGI = ∠BGH * 2.

Angle ABC can be represented as the sum of two angles: ∠ABC = ∠GBH + ∠GBI. Since the component angles are equal to each other (since they are formed by a bisector), angle ABC can be represented as the product of one of them and the number 2: ∠ABC = ∠GBH * 2.

Angles BGH and GBH are acute angles of a right triangle and therefore add up to 90°. Let's look at the resulting equalities:

∠BGH + ∠GBH = 90°
∠HGI = ∠BGH * 2
∠ABC = ∠GBH * 2

Let's add the last two:

∠HGI + ∠ABC = ∠BGH * 2 + ∠GBH * 2

Let's take the common factor out of brackets:

∠HGI + ∠ABC = 2(∠BGH + ∠GBH)

Since the sum of the angles in brackets is 90°, it turns out that angles HGI and ABC add up to 180°:

∠ABC + ∠HGI = 2 * 90° = 180°

So, we have proven that the sum of the angles HGI and ABC is 180°. Now let's look at the drawing again and return our attention to the angle with which angle ABC has correspondingly perpendicular sides. This is the DEF angle.

Lines GI and EF are parallel to each other since they are both perpendicular to the same line BC. And as you know, lines that are perpendicular to the same line are parallel to each other. For the same reason DE || GH.

As has been previously proven, angles with correspondingly parallel sides either add up to 180° or are equal to each other. This means either ∠DEF = ∠HGI, or ∠DEF + ∠HGI = 180°.

However, ∠ABC + ∠HGI = 180°. From this it is concluded that in the case of correspondingly perpendicular sides, the angles are either equal or add up to 180°.

Although in in this case we limited ourselves to proving only the amount. But if we mentally extend side EF in the opposite direction, we will see an angle that is equal to angle ABC, and at the same time its sides are also perpendicular to angle ABC. The equality of such angles can be proven by considering angles with correspondingly parallel sides: ∠DEF and ∠HGI.

53.Angles ( internal corners) triangle three angles are called, each of which is formed by three rays emerging from the vertices of the triangle and passing through the other two vertices.

54. Triangle Angle Sum Theorem. The sum of the angles of a triangle is 180°.

55. External corner of a triangle is an angle adjacent to some angle of this triangle.

56. External corner of a triangle is equal to the sum of two angles of a triangle that are not adjacent to it.

57. If all three corners triangle spicy, then the triangle is called acute-angled.

58. If one of the corners triangle blunt, then the triangle is called obtuse-angled.

59. If one of the corners triangle straight, then the triangle is called rectangular.

60. The side of a right triangle lying opposite the right angle is called hypotenuse(Greek word gyipotenusa - “contracting”), and two sides forming a right angle - legs(Latin word katetos - “plumb”) .

61. Theorem on the relationships between the sides and angles of a triangle. In a triangle against larger side there is a larger angle and back, The larger side lies opposite the larger angle.

62. In a right triangle The hypotenuse is longer than the leg.

because The larger side always lies opposite the larger angle.

Signs of an isosceles triangle.

If in a triangle two angles are equal, then it is isosceles;

If in a triangle the bisector is the median or height,
then this triangle is isosceles;

If in a triangle the median is the bisector or height, That

this triangle is isosceles;

If in a triangle height is median or bisector,

then this triangle is isosceles.

64. Theorem. Triangle inequality. The length of each side of the triangle is greater than the difference and less than the amount lengths of the other two sides:

Properties of the angles of a right triangle.

The sum of two acute angles of a right triangle is 90°.

A + B = 90°

66. Right Triangle Property.

A leg of a right triangle lying opposite an angle of 30° is equal to half the hypotenuse.

If/ A = 30°, then BC = ½ AB

67. Properties of a right triangle.

a) If a leg of a right triangle is equal to half the hypotenuse, then the angle opposite this leg is 30°.

If BC = ½ AB, then / B = 30°

B) The median drawn to the hypotenuse is equal to half the hypotenuse.

median CF = ½ AB

Sign of equality of right triangles on two sides.

If the legs of one right triangle are correspondingly equal to the legs of another, then such triangles are congruent.

For angles with correspondingly parallel sides, the following propositions are valid:

1. If sides a and b of one angle are respectively parallel to sides a and b of another angle and have the same directions as them, then the angles are equal.

2. If, under the same condition of parallelism, sides a and b are adjusted opposite to sides a and b, then the angles are also equal.

3. If, finally, the sides a and are parallel and identically directed, and the sides are parallel and oppositely directed, then the angles complement each other until they are reversed.

Proof. Let us prove the first of these propositions. Let the sides of the angles be parallel and equally directed (Fig. 191). Let's connect the vertices of the corners with a straight line.

In this case, two cases are possible: the straight line passes inside the corners or outside these corners (Fig. 191, b). In both cases the proof is obvious: so, in the first case

but where do we get it from? In the second case we have

and the result again follows from the equalities

We leave the proofs of Propositions 2 and 3 to the reader. We can say that if the sides of the angles are respectively parallel, then the angles are either equal or add up to the opposite angle.

Obviously, they are equal if both are simultaneously acute or both are obtuse, and their sum is equal if one of them is acute and the other is obtuse.

Angles with correspondingly perpendicular sides are equal or complementary to each other up to a straight angle.

Proof. Let a be some angle (Fig. 192), and O be the vertex of the angle formed by straight lines, respectively, it can be any of the four angles formed by these two straight lines). Let us rotate the angle (i.e., both of its sides) around its vertex O at a right angle; we obtain an angle equal to it, but one whose sides are perpendicular to the sides of the rotated angle indicated in Fig. 192 through They are parallel to straight lines forming given angle A. Therefore, angles mean that angles are either equal or form a reverse angle in total.