Glossary of planimetry terms- Definitions of terms from planimetry are collected here. References to terms in this glossary (on this page) are in italics. # A B C D E E E F G H I J K L M N O P R S ... Wikipedia

Collinear points

Competitive direct- Definitions of terms from planimetry are collected here. References to terms in this glossary (on this page) are in italics. # A B C D E E F G H I J K L M N O P R S T U V ... Wikipedia

Apollonia Circle- Definitions of terms from planimetry are collected here. References to terms in this glossary (on this page) are in italics. # A B C D E E F G H I J K L M N O P R S T U V ... Wikipedia

Plane transformation- Definitions of terms from planimetry are collected here. References to terms in this glossary (on this page) are in italics. # A B C D E E F G H I J K L M N O P R S T U V ... Wikipedia

Ceviana- Definitions of terms from planimetry are collected here. References to terms in this glossary (on this page) are in italics. # A B C D E E F G H I J K L M N O P R S T U V ... Wikipedia

Glossary of planimetry- This page is a glossary. See also the main article: Planimetry Definitions of terms from planimetry are collected here. Links to terms in this dictionary (on this page) are in italics... Wikipedia

Apollonius' problem- Apollonius' problem is to construct a circle tangent to three given circles using a compass and a ruler. According to legend, the problem was formulated by Apollonius of Perga around 220 BC. e. in the book “Touch”, which was lost ... Wikipedia

Apollonius' problem- Apollonius' problem is to construct a circle tangent to three given circles using a compass and a ruler. According to legend, the problem was formulated by Apollonius of Perga around 220 BC. e. in the book "Touching", which was lost, but was... ... Wikipedia

Voronoi diagram- a random set of points on the plane. The Voronoi diagram of a finite set of points S on the plane represents a partition of the plane such that ... Wikipedia

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

Theorem:

Bisectors AA 1, BB 1, СС 1 of the triangle intersect at one point O (Fig. 1).

Rice. 1. Illustration for the theorem

Proof:

Let us first consider two bisectors BB 1 and CC 1. They intersect, the intersection point O exists. To prove this, let us assume the opposite: let the given bisectors not intersect, in which case they are parallel. Then straight line BC is a secant and the sum of the angles is , this contradicts the fact that in the entire triangle the sum of the angles is .

So, point O of the intersection of two bisectors exists. Let's consider its properties:

Point O lies on the bisector of the angle, which means 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 - . Also, 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 obtained the following equalities:

, that is, all three perpendiculars dropped from 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 point O is equidistant from the sides of the angle, it follows that it lies on its bisector AA 1.

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

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

The segment AB is given. Any segment has a midpoint, and a perpendicular can be drawn through it - let’s denote it as 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 AO and OB are equal by condition, thus we have two right triangle, equal on two legs. It follows that the hypotenuses of the triangles are also equal, that is, what was required 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.

Given a segment AB, its perpendicular bisector p, and a point M equidistant from the ends of the segment. Prove that point M lies on the perpendicular bisector to the segment (Fig. 3).

Rice. 3. Illustration for the theorem

Proof:

Consider a triangle. It is isosceles, as per the condition. Consider the median of a triangle: point O is the middle of the base AB, OM is the median. According to the property of an isosceles triangle, the median drawn to its base is both an altitude and a bisector. It follows that . But line p is also perpendicular to AB. We know that at point O it is possible to draw a single perpendicular to the segment AB, which means the lines OM and p coincide, it follows that the point M belongs to the straight line p, which is what we needed to prove.

Direct and converse of the theorem 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, let us repeat that there are three segments in a triangle and the property of the perpendicular bisector applies to each of them.

Theorem:

The perpendicular bisectors of a triangle intersect at one point.

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

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

Rice. 4. Illustration for the theorem

Proof:

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

First level

Circumscribed circle. Visual Guide (2019)

The first question that may arise is: what is described - around what?

Well, actually, sometimes it happens around anything, but we will talk about a circle circumscribed around (sometimes they also say “about”) a triangle. What is it?

And just imagine, an amazing fact takes place:

Why is this fact surprising?

But triangles are different!

And for everyone there is a circle that will go through across all three peaks, that is, the circumscribed circle.

Proof of this amazing fact you can find in the following levels of the theory, but here we only note that if we take, for example, a quadrilateral, then not for everyone there will be a circle passing through the four vertices. For example, a parallelogram is an excellent quadrilateral, but there is no circle passing through all its four vertices!

And there is only for a rectangle:

Here you go, and every triangle always has its own circumscribed circle! And it’s even always quite easy to find the center of this circle.

Do you know what it is perpendicular bisector?

Now let's see what happens if we consider as many as three perpendicular bisectors to the sides of the triangle.

It turns out (and this is precisely what needs to be proven, although we will not) that all three perpendiculars intersect at one point. Look at the picture - all three perpendicular bisectors intersect at one point.

Do you think the center of the circumscribed circle always lies inside the triangle? Imagine - not always!

But if acute-angled, then - inside:

What to do with a right triangle?

And with an additional bonus:

Since we are talking about the radius of the circumscribed circle: what is it equal to for an arbitrary triangle? And there is an answer to this question: the so-called .

Namely:

And, of course,

1. Existence and circumcircle center

Here the question arises: does such a circle exist for every triangle? It turns out that yes, for everyone. And moreover, we will now formulate a theorem that also answers the question of where the center of the circumscribed circle is located.

Look like this:

Let's be brave and prove this theorem. If you have already read the topic “” and understood why three bisectors intersect at one point, then it will be easier for you, but if you haven’t read it, don’t worry: now we’ll figure it out.

We will carry out the proof using the concept of locus of points (GLP).

Well, for example, is the set of balls - “ locus» round objects? No, of course, because there are round...watermelons. Is it a set of people, a “geometric place”, who can speak? No, either, because there are babies who cannot speak. In life, it is generally difficult to find an example of a real “geometric location of points.” It's easier in geometry. Here, for example, is exactly what we need:

Here the set is the perpendicular bisector, and the property “ ” is “to be equidistant (a point) from the ends of the segment.”

Shall we check? So, you need to make sure of two things:

1. Any point that is equidistant from the ends of a segment is located on the perpendicular bisector to it.

Let's connect c and c.Then the line is the median and height b. This means - isosceles - we made sure that any point lying on the perpendicular bisector is equally distant from the points and.

Let's take the middle and connect and. The result is the median. But according to the condition, not only the median is isosceles, but also the height, that is, the perpendicular bisector. This means that the point exactly lies on the perpendicular bisector.

All! We have fully verified the fact that The perpendicular bisector of a segment is the locus of points equidistant from the ends of the segment.

This is all well and good, but have we forgotten about the circumscribed circle? Not at all, we have just prepared ourselves a “springboard for attack.”

Consider a triangle. Let's draw two bisectoral perpendiculars and, say, to the segments and. They will intersect at some point, which we will name.

Now, pay attention!

The point lies on the perpendicular bisector;
the point lies on the perpendicular bisector.
And that means, and.

Several things follow from this:

Firstly, the point must lie on the third bisector perpendicular to the segment.

That is, the perpendicular bisector must also pass through the point, and all three perpendicular bisectors intersect at one point.

Secondly: if we draw a circle with a center at a point and a radius, then this circle will also pass through both the point and the point, that is, it will be a circumscribed circle. This means that it already exists that the intersection of three perpendicular bisectors is the center of the circumscribed circle for any triangle.

And the last thing: about uniqueness. It is clear (almost) that the point can be obtained in a unique way, therefore the circle is unique. Well, we’ll leave “almost” for your reflection. So we proved the theorem. You can shout “Hurray!”

What if the problem asks “find the radius of the circumscribed circle”? Or vice versa, the radius is given, but you need to find something else? Is there a formula that relates the radius of the circumcircle to the other elements of the triangle?

Please note: the sine theorem states that in order to find the radius of the circumscribed circle, you need one side (any!) and the angle opposite to it. That's all!

3. Center of the circle - inside or outside

Now the question is: can the center of the circumscribed circle lie outside the triangle?
Answer: as much as possible. Moreover, this always happens in an obtuse triangle.

And generally speaking:

CIRCULAR CIRCLE. BRIEFLY ABOUT THE MAIN THINGS

1. Circle circumscribed about a triangle

This is the circle that passes through all three vertices of this triangle.

2. Existence and circumcircle center

Well, the topic is over. If you are reading these lines, it means you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!

Now the most important thing.

You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.

The problem is that this may not be enough...

For what?

For successful passing the Unified State Exam, for admission to college on a budget and, MOST IMPORTANTLY, for life.

I won’t convince you of anything, I’ll just say one thing...

People who received a good education, earn much more than those who did not receive it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...

But think for yourself...

What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?

GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.

You won't be asked for theory during the exam.

You will need solve problems against time.

And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.

It's like in sports - you need to repeat it many times to win for sure.

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Instructions

Draw a straight line through the intersection points of the circles. You have obtained the perpendicular bisector to a given segment.

Let us now be given a point and a straight line. It is necessary to draw a perpendicular from this point to. Place the needle at the point. Draw a circle of radius (the radius must be from a point to a line so that the circle can intersect the line at two points). Now you have two points on a line. These points create a line segment. Construct the perpendicular bisector to the segment, the ends are the resulting points, according to the algorithm discussed above. The perpendicular must pass through the starting point.

Construction of straight lines is the basis of technical drawing. Nowadays this is increasingly done with the help of graphic editors, which provide the designer with great opportunities. However, some principles of construction remain the same as in classical drawing - using a pencil and ruler.

You will need

• - paper;
• - pencil;
• - ruler;
• - computer with AutoCAD program.

Instructions

Start with the classic construction. Determine the plane in which you will build the line. Let this be the plane of a sheet of paper. Depending on the conditions of the problem, arrange. They can be arbitrary, but it is possible that a coordinate system is given. Place arbitrary points where you like best. Label them A and B. Use a ruler to connect them. According to the axiom, it is always possible to draw a straight line through two points, and only one.

Draw a coordinate system. Let you be given points A (x1; y1). To make them, you need to plot the required number along the x-axis and draw a straight line parallel to the y-axis through the marked point. Then plot the value equal to y1 along the corresponding axis. From the marked point, draw a perpendicular until it intersects with. The place of their intersection will be point A. In the same way, find point B, the coordinates of which can be designated as (x2; y2). Connect both points.

In AutoCAD, a straight line can be constructed using several . The "by" function is usually installed by default. Find the “Home” tab in the top menu. You will see the Draw panel in front of you. Find the button with the image of a straight line and click on it.

AutoCAD also allows you to specify the coordinates of both . Type (_xline) in the command line below. Press Enter. Enter the coordinates of the first point and also press enter. Determine the second point in the same way. It can also be specified by clicking the mouse, placing the cursor in desired point screen.

In AutoCAD, you can build a straight line not only by two points, but also by the angle of inclination. IN context menu"Draw" select the straight line and then the "Angle" option. The starting point can be set by clicking the mouse or by , as in the previous method. Then set the angle size and press enter. By default, the straight line will be located at the desired angle to the horizontal.

Video on the topic

On a complex drawing (diagram) perpendicularity straight and plane determined by the basic provisions: if one side right angle parallel plane projections, then a right angle is projected onto this plane without distortion; if a line is perpendicular to two intersecting lines plane, it is perpendicular to this plane.

You will need

• Pencil, ruler, protractor, triangle.

Instructions

Example: draw a perpendicular through point M to plane To draw a perpendicular to plane, there are two intersecting lines lying in this plane, and construct a line perpendicular to them. The frontal and horizontal are chosen as these two intersecting lines. plane.

Frontal f(f₁f₂) is a straight line lying in plane and parallel to the frontal plane projections P₂. This means f₂ is its natural value, and f₁ is always parallel to x₁₂. From point A₂, draw h₂ parallel to x₁₂ and get point 1₂ on B₂C₂.

Using a projection communication line, point 1₁ to B₁C₁. Connect with A₁ – this is h₁ – natural size horizontal. From point B₁ draw f₁‖x₁₂, at A₁C₁ you get point 2₁. Using the projection connection line, find point 2₂ on A₂C₂. Connect to point B₂ - this will be f₂ - the natural size of the front.

Constructed natural horizontals h₁ and frontals f₂ of projections perpendicular to plane. From point M₂, draw its frontal projection a₂ at an angle of 90

Perpendicular bisector (median perpendicular or mediatrix) - a straight line perpendicular to a given segment and passing through its middle.

Properties

$p\_a=\backslash tfrac(2aS)(a^2+b^2-c^2),\; p\_b=\backslash tfrac(2bS)(a^2+b^2-c^2),\; p\_c=\backslash tfrac(2cS)(\; a^2-b^2+c^2),$ where the subscript denotes 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 $a\backslash geqslant\; b\backslash geqslant\; c.$ $p\_a\backslash geq\; p\_b$ And $p\_c\backslash geq\; p\_b.$ In other words, the smallest perpendicular bisector of a triangle belongs to the middle segment.

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Notes

Excerpt characterizing the perpendicular bisector

Kutuzov, stopping to chew, stared at Wolzogen in surprise, as if not understanding what was being said to him. 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, frowning, quickly getting up and advancing on Wolzogen. “How do you... how dare you!..”, he shouted, making threatening gestures with shaking hands and choking. - How dare you, 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, 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 the next day my absolute intention to attack the enemy,” Kutuzov said sternly. Everyone was silent, and all that could be heard was the heavy breathing of the out of breath old general. “They were 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 sobbed from the tears that came. Wolzogen, shrugging his shoulders and pursing his lips, silently walked away to the side, wondering uber diese Eingenommenheit des alten Herrn. [at this tyranny of the old gentleman. (German) ]
“Yes, here he is, my hero,” Kutuzov said to the plump, handsome, black-haired general, who was entering the mound at that time. It was Raevsky, who 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? [You don't think, then, like others, that we should retreat?]