Definition of the mean perpendicular. Four wonderful points of the triangle

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

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 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 $a\; \backslash geqslant\; b\; \backslash geqslant\; c.$ $p\_a\backslash geq\; p\_b$ And $p\_c\backslash geq\; p\_b.$ 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 ...
- You saw? 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?]

Instruction

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

Now let us be given a point and a 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 the line. These points create a line. Construct the perpendicular bisector to the segment, the ends are the obtained points, according to the algorithm discussed above. The perpendicular must pass through the starting point.

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

You will need

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

Instruction

Start with a classic build. Determine the plane in which you will draw 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. Arbitrary points put 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). In order for them, it is necessary to set aside the required number along the x-axis and draw a straight line parallel to the y-axis through the marked point. Then plot a value equal to y1 along the corresponding axis. Draw a perpendicular from the marked point 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 denoted as (x2; y2). Connect both dots.

In AutoCAD, a straight line can be built with several . The "by" function is usually set by default. Find the "Home" tab in the top menu. You will see the Drawing panel in front of you. Find the button with the straight line and click on it.

AutoCAD also allows you to set the coordinates of both . Dial in the bottom command line(_xline). Press Enter. Enter the coordinates of the first point and also press enter. Define the second point in the same way. It can also be specified with a mouse click by 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. From the Draw context menu, select a straight line and then the Angle option. The starting point can be set by mouse click or by , as in the previous method. Then set the corner size and hit enter. By default, the line will be positioned at the desired angle to the horizontal.

Related videos

On a complex drawing (diagram) perpendicularity direct and plane determined by the main 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.

Instruction

Example: through point M draw a perpendicular 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 the horizontal are chosen as these two intersecting lines. plane.

The frontal f(f₁f₂) is a straight line lying in plane and parallel to the front plane projections П₂. So 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.

With the help of a projection line of communication point 1₁ on В₁С₁. Connect with A₁ - this is h₁ - the natural size of the horizontal. From point B₁ draw f₁‖x₁₂, on A₁C₁ get point 2₁. Find the point 2₂ on A₂C₂ using the projection connection line. Connect with point B₂ - this will be f₂ - the full size of the front.

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

There are so-called four remarkable points in a triangle: the point of intersection of the medians. The point of intersection of the bisectors, the point of intersection of the heights and the point of intersection of the perpendicular bisectors. Let's consider each of them.

Point of intersection of the medians of a triangle

Theorem 1

On the intersection of the medians of a triangle: The medians of a triangle intersect at one point and divide the intersection point in a ratio of $2:1$ starting from the vertex.

Proof.

Consider triangle $ABC$, where $(AA)_1,\ (BB)_1,\ (CC)_1$ is its median. Since the medians divide the sides in half. Consider the middle line $A_1B_1$ (Fig. 1).

Figure 1. Medians of a triangle

By Theorem 1, $AB||A_1B_1$ and $AB=2A_1B_1$, hence $\angle ABB_1=\angle BB_1A_1,\ \angle BAA_1=\angle AA_1B_1$. Hence the triangles $ABM$ and $A_1B_1M$ are similar in the first similarity triangles. Then

Similarly, it is proved that

The theorem has been proven.

Intersection point of the bisectors of a triangle

Theorem 2

On the intersection of the bisectors of a triangle: The bisectors of a triangle intersect at one point.

Proof.

Consider triangle $ABC$, where $AM,\ BP,\ CK$ are its bisectors. Let the point $O$ be the intersection point of the bisectors $AM\ and\ BP$. Draw from this point perpendicular to the sides of the triangle (Fig. 2).

Figure 2. Bisectors of a triangle

Theorem 3

Each point of the bisector of a non-expanded angle is equidistant from its sides.

By Theorem 3, we have: $OX=OZ,\ OX=OY$. Hence $OY=OZ$. Hence the point $O$ is equidistant from the sides of the angle $ACB$ and therefore lies on its bisector $CK$.

The theorem has been proven.

Intersection point of the perpendicular bisectors of a triangle

Theorem 4

The perpendicular bisectors of the sides of a triangle intersect at one point.

Proof.

Let a triangle $ABC$ be given, $n,\ m,\ p$ its perpendicular bisectors. Let the point $O$ be the intersection point of the perpendicular bisectors $n\ and\ m$ (Fig. 3).

Figure 3. Perpendicular bisectors of a triangle

For the proof we need the following theorem.

Theorem 5

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

By Theorem 3, we have: $OB=OC,\ OB=OA$. Hence $OA=OC$. This means that the point $O$ is equidistant from the ends of the segment $AC$ and, therefore, lies on its perpendicular bisector $p$.

The theorem has been proven.

The point of intersection of the altitudes of the triangle

Theorem 6

The heights of a triangle or their extensions intersect at one point.

Proof.

Consider triangle $ABC$, where $(AA)_1,\ (BB)_1,\ (CC)_1$ is its height. Draw a line through each vertex of the triangle parallel to the side opposite the vertex. We get a new triangle $A_2B_2C_2$ (Fig. 4).

Figure 4. Heights of a triangle

Since $AC_2BC$ and $B_2ABC$ are parallelograms with a common side, then $AC_2=AB_2$, that is, point $A$ is the midpoint of side $C_2B_2$. Similarly, we get that the point $B$ is the midpoint of the side $C_2A_2$, and the point $C$ is the midpoint of the side $A_2B_2$. From the construction we have that $(CC)_1\bot A_2B_2,\ (BB)_1\bot A_2C_2,\ (AA)_1\bot C_2B_2$. Hence $(AA)_1,\ (BB)_1,\ (CC)_1$ are the perpendicular bisectors of triangle $A_2B_2C_2$. Then, by Theorem 4, we have that the heights $(AA)_1,\ (BB)_1,\ (CC)_1$ intersect at one point.