Midperpendicular in a right triangle. A circle circumscribed about a triangle. A triangle inscribed in a circle

Proofs of theorems on the properties of a circle circumscribed about a triangle

Midperpendicular to the segment

Definition 1 . Midperpendicular to the segment called, a straight line perpendicular to this segment and passing through its middle (Fig. 1).

Theorem 1. Each point of the perpendicular bisector to the segment is at the same distance from the ends this segment.

Proof . Consider an arbitrary point D lying on the perpendicular bisector to the segment AB (Fig. 2), and prove that triangles ADC and BDC are equal.

Indeed, these triangles are right-angled triangles whose legs AC and BC are equal, while the legs DC are common. From the equality of triangles ADC and BDC, the equality of segments AD and DB follows. Theorem 1 is proved.

Theorem 2 (Reverse to Theorem 1). If a point is at the same distance from the ends of a segment, then it lies on the perpendicular bisector to this segment.

Proof . Let us prove Theorem 2 by the method “by contradiction”. To this end, suppose that some point E is at the same distance from the ends of the segment, but does not lie on the perpendicular bisector to this segment. Let us bring this assumption to a contradiction. Let us first consider the case when points E and A lie on opposite sides of the perpendicular bisector (Fig. 3). In this case, the segment EA intersects the perpendicular bisector at some point, which we will denote by the letter D.

Let us prove that the segment AE is longer than the segment EB . Really,

Thus, in the case when the points E and A lie on opposite sides of the perpendicular bisector, we have obtained a contradiction.

Now consider the case when points E and A lie on the same side of the perpendicular bisector (Fig. 4). Let us prove that the segment EB is longer than the segment AE . Really,

The resulting contradiction completes the proof of Theorem 2

Circle circumscribing a triangle

Definition 2 . A circle circumscribing a triangle, call the circle passing through all three vertices of the triangle (Fig. 5). In this case the triangle is called a triangle inscribed in a circle or inscribed triangle.

Properties of a circle circumscribed about a triangle. Sine theorem

Figure	Drawing	Property
Midperpendiculars to the sides of the triangle		intersect at one point .

Center circumscribed about an acute triangle of a circle		Center described about acute-angled inside triangle.
Center circle circumscribed about a right triangle		The center of the described about rectangular midpoint of the hypotenuse .
Center circumscribed about an obtuse triangle of a circle		Center described about obtuse circle triangle lies outside triangle.
		,
Square triangle		S= 2R 2 sin A sin B sin C ,
Radius of the circumscribed circle		For any triangle, the equality is true:

Midperpendiculars to the sides of a triangle

All perpendicular bisectors drawn to the sides of an arbitrary triangle, intersect at one point .

Circle circumscribing a triangle

Any triangle can be circumscribed by a circle. . The center of the circle circumscribed about the triangle is the point where all the perpendicular bisectors drawn to the sides of the triangle intersect.

Center of a circle circumscribed about an acute triangle

Center described about acute-angled circle triangle lies inside triangle.

Center of a circle circumscribed about a right triangle

The center of the described about rectangular circle triangle is midpoint of the hypotenuse .

Center of a circle circumscribed about an obtuse triangle

Center described about obtuse circle triangle lies outside triangle.

For any triangle, equalities are valid (sine theorem):

where a, b, c are the sides of the triangle, A, B, C are the angles of the triangle, R is the radius of the circumscribed circle.

Area of a triangle

For any triangle, the equality is true:

S= 2R 2 sin A sin B sin C ,

where A, B, C are the angles of the triangle, S is the area of the triangle, R is the radius of the circumscribed circle.

Radius of the circumscribed circle

For any triangle, the equality is true:

where a, b, c are the sides of the triangle, S is the area of the triangle, R is the radius of the circumscribed circle.

Proofs of theorems on the properties of a circle circumscribed about a triangle

Theorem 3. All midperpendiculars drawn to the sides of an arbitrary triangle intersect at one point.

Proof . Consider two perpendicular bisectors drawn to the sides AC and AB of the triangle ABC , and denote the point of their intersection with the letter O (Fig. 6).

Since the point O lies on the perpendicular bisector to the segment AC , then by virtue of Theorem 1 the equality holds.

Midperpendicular (median perpendicular or mediatrix) is a straight line perpendicular to the given segment and passing through its midpoint.

Properties

p_a=\tfrac(2aS)(a^2+b^2-c^2), p_b=\tfrac(2bS)(a^2+b^2-c^2), p_c=\tfrac(2cS)(  a^2-b^2+c^2),

where the subscript indicates the side to which the perpendicular is drawn,

S

is the area of the triangle, and it is also assumed that the sides are related by inequalities

a \geqslant b \geqslant c.

p_a\geq p_b

And

p_c\geq p_b.

In other words, for a triangle, the smallest perpendicular bisector refers to the middle segment.

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Notes

An excerpt characterizing the perpendicular bisector

Kutuzov, stopping to chew, stared at Wolzogen in surprise, as if not understanding what he was being told. Wolzogen, noticing the excitement of des alten Herrn, [the old gentleman (German)], said with a smile:
- I did not consider myself entitled to hide from your lordship what I saw ... The troops are in complete disorder ...
- Have you seen? Did you see? .. - Kutuzov shouted with a frown, quickly getting up and advancing on Wolzogen. “How dare you… how dare you…!” he shouted, making menacing gestures with shaking hands and choking. - How dare you, my dear sir, say this to me. You don't know anything. Tell General Barclay from me that his information is incorrect and that the real course of the battle is known to me, the commander-in-chief, better than to him.
Wolzogen wanted to object something, but Kutuzov interrupted him.
- The enemy is repulsed on the left and defeated on the right flank. If you have not seen well, dear sir, then do not allow yourself to say what you do not know. Please go to General Barclay and convey to him my indispensable intention to attack the enemy tomorrow, ”Kutuzov said sternly. Everyone was silent, and one could hear one heavy breathing of the out of breath old general. - Repulsed everywhere, for which I thank God and our brave army. The enemy is defeated, and tomorrow we will drive him out of the sacred Russian land, - said Kutuzov, crossing himself; and suddenly burst into tears. Wolzogen, shrugging his shoulders and twisting his lips, silently stepped aside, wondering at uber diese Eingenommenheit des alten Herrn. [on this tyranny of the old gentleman. (German)]
“Yes, here he is, my hero,” Kutuzov said to the plump, handsome black-haired general, who at that time was entering the mound. It was Raevsky, who had spent the whole day at the main point of the Borodino field.
Raevsky reported that the troops were firmly in their places and that the French did not dare to attack anymore. After listening to him, Kutuzov said in French:
– Vous ne pensez donc pas comme lesautres que nous sommes obliges de nous retirer? [So you don't think, like the others, that we should retreat?]

To give an idea of a new class of problems - the construction of geometric figures using a compass and a ruler without scale divisions.

Introduce the concept of GMT.

Give a definition of the perpendicular bisector, teach how to build it and prove the term about the perpendicular bisector, as well as its inverse.

Using the Compass-3D computer drawing system, perform geometric constructions, which are recommended to be carried out in a geometry course using a compass and ruler.

Handout (Appendix No. 1)

Problems for building with a compass and a ruler without divisions are most often solved according to a certain scheme:

I. Analysis: Draw the desired figure schematically and establish links between the problem data and the desired elements.

II. Building: According to the plan, they build with a compass and a ruler.

III. Proof: Prove that the constructed figure satisfies the conditions of the problem.

IV. Study: Conduct a study, for any data, whether the problem has a solution and if so, how many solutions (do not perform in all problems).

Here are some examples of elementary construction tasks that we will consider:

1. Set aside a segment equal to this one (studied earlier).

2. Construction of the perpendicular bisector to the segment:

construct the midpoint of the given segment;
construct a line passing through a given point and perpendicular to a given line (a point may or may not lie on a given line).

3. Construction of the angle bisector.

4. Construction of an angle equal to a given one.

The median perpendicular to the segment.

Definition: The perpendicular bisector of a segment is a line passing through the midpoint of the segment and perpendicular to it.

Task: "Construct the perpendicular bisector to the segment." Presentation

O - the middle of AB

Construction description ( slide number 4):

Beam a; A - the beginning of the beam

Circumference (A; r =m)

Circle a = B; AB = m

Circle 1 (A; r 1 > m/2)

Circle 2 (B; r 1)

Circle 1 Circle 2 =

MN ; MN AB =0, (MN = L)

where MN AB, O is the midpoint of AB

III. Proof(slide number 5, 6)

1. Consider AMN and BNM:

AM = MB=BN=AN=r 2 , therefore AM = BN , AN = BM MN is the common side

(Figure 3)

Therefore, AMN = BNM (on 3 sides),

Hence

1= 2 (by definition equal)

3= 4 (by definition equal)

2. MAN and NBM are isosceles (by definition) ->

1 \u003d 4 and 3 \u003d 2 (by the property of isosceles)

3. From points 1 and 2 -> 1 = 3 therefore MO is the bisector of the isosceles AMB

4. Thus we have proved that MN is the perpendicular bisector to the segment AB

IV. Study

This problem has a unique solution, because Any line segment has only one midpoint, and through a given point one can draw a single line perpendicular to the given one.

Definition: A geometric set of points (GMT) is a set of points that have some property. (Appendix No. 2)

Known to you GMT:

The perpendicular bisector of a segment is the set of points equidistant from the ends of the segment.
Bisector of an angle - a set of points equidistant from the sides of the angle

So let's prove the theorem:

Theorem: "Each point of the perpendicular bisector to a segment is equidistant from the ends of this segment."

(Figure 4)

Given: AB; MO - perpendicular bisector

Prove: AM = VM

Proof:

1. MO - perpendicular bisector (by condition) -> O - midpoint of segment AB, MOAB

2. Consider AMO and WMO - rectangular

MO - common leg

AO \u003d VO (O - the middle of AB) -\u003e AMO \u003d BMO (on 2 legs) -\u003e AM \u003d VM (by definition of equal triangles, as corresponding sides)

Q.E.D

Homework: “Prove the theorem inverse to the given one”

Theorem: “Each point equidistant from the ends of a segment lies on the perpendicular bisector to this segment.”

(Figure 5)

Given: AB; MA=MV

Prove: Point M lies on the perpendicular bisector

Proof:

That. MO - perpendicular bisector containing all points equidistant from the ends of the segment.

Property of perpendicular bisectors to the sides of a triangle

They intersect at one point and this point is the center of the circumcircle around the triangle, we will study in the eighth grade.

Workshop

Material and technical equipment:

Distribution: 29,574 KB

OS: Windows 9x/2000/XP

Website: http://www.ascon.ru

Now we will transfer the construction to the graphical environment of the computer (slide number 7)

Previously acquired knowledge and skills must be applied to a specific task. You will see that the construction will take you no more time than the construction in a notebook. Among other things, it is interesting to see how the computer environment executes human commands to build planar figures. Before you is appendix No. 3, in which your construction steps are described in detail. Load the program and open a new drawing ( slide number 8, 9).

Draw geometric objects specified in the problem condition: ray A starting at point A and the segment is equal m– arbitrary length ( slide number 10).

Enter the designation of the beam, segment, beginning of the beam in the drawing using the tab "Tools" text.

Construct a circle with a radius equal to the segment m centered at the vertex by a given point A (slide number 11).

m centered at the vertex given point A ( slide №12, 13).

Construct a circle with a radius equal to a segment greater than 1/2 m To do this, select the item “ Between 2 points” (slide №14, 15, 16).

Through the intersection points of the circles M and N draw a line ( slide №17,18).

Used Books:

Ugrinovich N.D. “Informatics. Basic course” Grade 7. - M.: BINOM - 2008 - 175 p.
Ugrinovich N.D. “Workshop on informatics and information technologies”. Tutorial. - M.: BINOM, 2004-2006. -
Ugrinovich N.D. “Teaching the course “Informatics and ICT” in elementary and high school grades 8-11 M.: BINOM Knowledge Laboratory, 2008. - 180 p.
Ugrinovich ND Computer workshop on CD-ROM. - M.: BINOM, 2004-2006.
Boguslavsky A.A., Tretyak T.M. Farafonov A.A. “Compass - 3D v 5.11-8.0 Workshop for beginners” - M .: SOLON - PRESS, 2006 - 272 p.
Atanasyan L.S., Butuzov V.F., Kadomtsev S.B., et al “Geometry 7-9. Textbook for secondary schools "- M: Education 2006 - 384 p.
Atanasyan L.S., Butuzov V.F., Kadomtsev S.B., et al “Study of geometry grades 7-9. Guidelines for the textbook "- M: Education 1997 - 255 p.
Afanas'eva T.L., Tapilina L.A. “Lesson plans for the 8th grade textbook of Atanasyan L.S.” - Volgograd "Teacher" 2010, 166 p.

Application No. 1

Plan for solving problems on the construction of a compass and a ruler.

Analysis.
Construction.
Proof.
Study.

Explanation

When performing the analysis, the required figure is schematically drawn and a connection is established between the task data and the required elements.
According to the plan, the construction is carried out with a compass and a ruler.
They prove that the constructed figure satisfies the conditions of the problem.
Conduct a study: for any data, does the problem have a solution, and if so, how many solutions?

Examples of elementary construction tasks

Set aside a segment equal to the given one.
Construct a perpendicular bisector to a segment.
Construct the midpoint of the segment.
Construct a line passing through the given point, perpendicular to the given line (The point may or may not lie on the given line).
Construct an angle bisector.
Construct an angle equal to the given one.

Application №2

The locus of points (GMT) is a set of points that have some property.

Examples of GMT:

The perpendicular bisector of a segment is the set of points equidistant from the ends of the segment.
A circle is a set of points equidistant from a given point - the center of the circle.
The bisector of an angle is the set of points equidistant from the sides of the angle.

Each point of the perpendicular bisector to a segment is equidistant from the ends of this segment.

In the previous lesson, we considered the properties of the bisector of an angle, both enclosed in a triangle and free. The triangle includes three angles, and for each of them the considered properties of the bisector are preserved.

Theorem:

The bisectors AA 1, BB 1, CC 1 of the triangle intersect at one point O (Fig. 1).

Rice. 1. Illustration for the theorem

Proof:

Consider first two bisectors BB 1 and СС 1 . They intersect, the intersection point O exists. To prove this, suppose the contrary: let the given bisectors do not intersect, in which case they are parallel. Then the line BC is a secant and the sum of the angles , this contradicts the fact that in the whole triangle the sum of the angles is .

So, the point O of intersection of two bisectors exists. Consider its properties:

Point O lies on the bisector of angle , which means that it is equidistant from its sides BA and BC. If OK is perpendicular to BC, OL is perpendicular to BA, then the lengths of these perpendiculars are equal to -. Also, the point O lies on the bisector of the angle and is equidistant from its sides CB and CA, the perpendiculars OM and OK are equal.

We got the following equalities:

, that is, all three perpendiculars dropped from the point O to the sides of the triangle are equal to each other.

We are interested in the equality of perpendiculars OL and OM. This equality says that the point O is equidistant from the sides of the angle, hence it lies on its bisector AA 1.

Thus, we have proved that all three bisectors of a triangle intersect at one point.

In addition, the triangle consists of three segments, which means that we should consider the properties of a single segment.

Segment AB is given. Any segment has a middle, and a perpendicular can be drawn through it - we denote it by p. Thus p is the perpendicular bisector.

Rice. 2. Illustration for the theorem

Any point lying on the perpendicular bisector is equidistant from the ends of the segment.

Prove that (Fig. 2).

Proof:

Consider triangles and . They are rectangular and equal, because they have a common leg OM, and the legs of AO and OB are equal by condition, so we have two right-angled triangles that are equal in two legs. It follows that the hypotenuses of the triangles are also equal, that is, which was to be proved.

The converse theorem is true.

Each point equidistant from the ends of a segment lies on the perpendicular bisector to this segment.

The segment AB is given, the perpendicular bisector to it is p, the point M is equidistant from the ends of the segment. Prove that the point M lies on the perpendicular bisector to the segment (Fig. 3).

Rice. 3. Illustration for the theorem

Proof:

Let's consider a triangle. It is isosceles, as by condition. Consider the median of the triangle: point O is the midpoint of the base AB, OM is the median. According to the property of an isosceles triangle, the median drawn to its base is both a height and a bisector. Hence it follows that . But the line p is also perpendicular to AB. We know that a single perpendicular to the segment AB can be drawn to the point O, which means that the lines OM and p coincide, hence it follows that the point M belongs to the line p, which was required to be proved.

The direct and inverse theorems can be generalized.

A point lies on the perpendicular bisector of a segment if and only if it is equidistant from the ends of this segment.

So, we repeat that there are three segments in a triangle and the property of the perpendicular bisector is applicable to each of them.

Theorem:

The perpendicular bisectors of a triangle intersect at one point.

A triangle is given. Perpendicular to its sides: P 1 to side BC, P 2 to side AC, P 3 to side AB.

Prove that the perpendiculars Р 1 , Р 2 and Р 3 intersect at the point O (Fig. 4).

Rice. 4. Illustration for the theorem

Proof:

Consider two midperpendiculars P 2 and P 3 , they intersect, the intersection point O exists. Let us prove this fact by contradiction - let the perpendiculars P 2 and P 3 be parallel. Then the angle is straight, which contradicts the fact that the sum of the three angles of a triangle is . So, there is a point O of intersection of two of the three perpendicular bisectors. Properties of the point O: it lies on the perpendicular bisector to the side AB, which means that it is equidistant from the ends of the segment AB:. It also lies on the perpendicular bisector to side AC, so . We have obtained the following equalities.