Formulate the property of a tangent to a circle. Tangent line. Theorem on arcs contracted by equal chords

Definition. A tangent to a circle is a straight line in the plane that has exactly one common point with the circle.

Here are a couple of examples:

Circle with center O touches a straight line l at the point A

From anywhere M Exactly two tangents can be drawn outside the circle

difference between tangent l, secant BC and direct m, which has no common points with the circle

This could be the end, but practice shows that it is not enough just to memorize the definition - you need to learn to see the tangents in the drawings, know their properties and, in addition, how to practice using these properties when solving real problems. We will deal with all this today.

Basic properties of tangents

In order to solve any problems, you need to know four key properties. Two of them are described in any reference book / textbook, but the last two are somehow forgotten about, but in vain.

1. Segments of tangents drawn from one point are equal

A little higher, we already talked about two tangents drawn from one point M. So:

The segments of the tangents to the circle, drawn from one point, are equal.

Segments AM and BM equal

2. The tangent is perpendicular to the radius drawn to the point of contact

Let's look at the picture above again. Let's draw the radii OA and OB, after which we find that the angles OAM and OBM- straight.

The radius drawn to the tangent point is perpendicular to the tangent.

This fact can be used without proof in any problem:

The radii drawn to the tangent point are perpendicular to the tangents

By the way, note: if you draw a segment OM, then we get two equal triangles: OAM and OBM.

3. Relationship between tangent and secant

But this is a more serious fact, and most schoolchildren do not know it. Consider a tangent and a secant that pass through the same common point M. Naturally, the secant will give us two segments: inside the circle (segment BC- it is also called a chord) and outside (it is called that - the outer part MC).

The product of the entire secant by its outer part is equal to the square of the tangent segment

Relationship between secant and tangent

4. Angle between tangent and chord

An even more advanced fact that is often used to solve complex problems. I highly recommend taking it on board.

The angle between a tangent and a chord is equal to the inscribed angle based on this chord.

Where does the dot come from B? In real problems, it usually "pops up" somewhere in the condition. Therefore, it is important to learn to recognize this configuration in the drawings.

Sometimes it still applies :)

The concept of a tangent to a circle

The circle has three possible mutual positions relative to the straight line:

If the distance from the center of the circle to the line is less than the radius, then the line has two points of intersection with the circle.

If the distance from the center of the circle to the line is equal to the radius, then the line has two points of intersection with the circle.

If the distance from the center of the circle to the straight line is greater than the radius, then the straight line has two points of intersection with the circle.

We now introduce the concept of a tangent line to a circle.

Definition 1

A tangent to a circle is a straight line that has one point of intersection with it.

The common point of the circle and the tangent is called the tangent point (Fig. 1).

Figure 1. Tangent to a circle

Theorems related to the concept of a tangent to a circle

Theorem 1

Tangent property theorem: The tangent to the circle is perpendicular to the radius drawn to the tangent point.

Proof.

Consider a circle with center $O$. Let us draw the tangent $a$ at the point $A$. $OA=r$ (Fig. 2).

Let us prove that $a\bot r$

We will prove the theorem by the method of "by contradiction". Assume that the tangent $a$ is not perpendicular to the radius of the circle.

Figure 2. Illustration of Theorem 1

That is, $OA$ is oblique to a tangent. Since the perpendicular to the line $a$ is always less than the slope to the same line, the distance from the center of the circle to the line is less than the radius. As we know, in this case the line has two points of intersection with the circle. Which contradicts the definition of a tangent.

Therefore, the tangent is perpendicular to the radius of the circle.

The theorem has been proven.

Theorem 2

Converse to the tangent property theorem: If the line passing through the end of the radius of a circle is perpendicular to the radius, then this line is tangent to this circle.

Proof.

According to the condition of the problem, we have that the radius is a perpendicular drawn from the center of the circle to the given line. Therefore, the distance from the center of the circle to the straight line is equal to the length of the radius. As we know, in this case the circle has only one point of intersection with this line. By definition 1, we get that the given line is tangent to the circle.

The theorem has been proven.

Theorem 3

The segments of the tangents to the circle, drawn from one point, are equal and make equal angles with the line passing through this point and the center of the circle.

Proof.

Let there be given a circle centered at the point $O$. Two different tangents are drawn from the point $A$ (which lies on all circles). From the touch point $B$ and $C$ respectively (Fig. 3).

Let us prove that $\angle BAO=\angle CAO$ and that $AB=AC$.

Figure 3. Illustration of Theorem 3

By Theorem 1, we have:

Therefore, the triangles $ABO$ and $ACO$ are right triangles. Since $OB=OC=r$, and the hypotenuse $OA$ is common, these triangles are equal in hypotenuse and leg.

Hence we get that $\angle BAO=\angle CAO$ and $AB=AC$.

The theorem has been proven.

An example of a task on the concept of a tangent to a circle

Example 1

Given a circle with center $O$ and radius $r=3\ cm$. The tangent $AC$ has a tangent point $C$. $AO=4\cm$. Find $AC$.

Solution.

First, let's depict everything in the figure (Fig. 4).

Figure 4

Since $AC$ is a tangent and $OC$ is a radius, then by Theorem 1 we get $\angle ACO=(90)^(()^\circ )$. It turned out that the triangle $ACO$ is rectangular, which means, according to the Pythagorean theorem, we have:

\[(AC)^2=(AO)^2+r^2\] \[(AC)^2=16+9\] \[(AC)^2=25\] \

\[(\Large(\text(Central and Inscribed Angles)))\]

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 the circle.

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

Theorem

The measure of an inscribed angle is half the measure of the arc it intercepts.

Proof

We will carry out the proof in two stages: first, we prove the validity of the statement for the case when one of the sides of the inscribed angle contains a diameter. Let the 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 outer, 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 $ABC$ . Draw the circle diameter $BD$ from the vertex of the inscribed angle. Two cases are possible:

1) the diameter cut the angle into two angles $\angle ABD, \angle CBD$ (for each of which the theorem is true as proved above, therefore it is also true for the original angle, which is the sum of these two and, therefore, is equal to half the sum of the arcs on which they lean, that is, equal to half of the arc on which it leans). Rice. one.

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, it is equal to half the arc on which it rests). Rice. 2.

Consequences

1. Inscribed angles based on the same arc are equal.

2. An inscribed angle based on a semicircle is a right angle.

3. An inscribed angle is equal to half the central angle based on the same arc.

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

Definitions

There are three types relative position straight line and circle:

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

Theorem

1. The tangent to the circle is perpendicular to the radius drawn to the point of contact.

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

Consequence

The segments of tangents drawn from one point to the circle are equal.

Proof

Draw two tangents $KA$ and $KB$ to the circle from the point $K$:

So $OA\perp KA, OB\perp KB$ as radii. Right triangles $\triangle KAO$ and $\triangle KBO$ are equal in leg and hypotenuse, hence $KA=KB$ .

Consequence

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

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

The theorem about the angle between secants

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

Proof

Let $M$ be a point from which two secants are drawn as shown in the figure:

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

$\angle DAB$ is the outer corner 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 to be proved.

Angle theorem 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 $\angle AMD = 180^\circ - \angle CMD$, whence 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 tangent point is equal to half the degree measure of the arc subtracted by the chord.

Proof

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

Denote $\angle OAB = \alpha$ . Since $OA$ and $OB$ are radii, then $OA = OB$ and $\angle OBA = \angle OAB = \alpha$. In this way, $\buildrel\smile\over(AB) = \angle AOB = 180^\circ - 2\alpha = 2(90^\circ - \alpha)$.

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

Theorem on arcs contracted by equal chords

Equal chords subtend equal arcs, smaller semicircles.

And vice versa: equal arcs are contracted 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 since $\angle AOB, \angle COD$ - central corners based on arcs $\buildrel\smile\over(AB), \buildrel\smile\over(CD)$ respectively, then $\buildrel\smile\over(AB)=\buildrel\smile\over(CD)$.

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

Theorem

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

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

Proof

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

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

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

Similarly, $\triangle AOB$ is isosceles, $ON$ is the height, so $ON$ is the median. Therefore, $AN=NB$ .

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

Theorem on the product of segments of chords

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, the angles $1$ and $2$ are equal, since they are inscribed and rely on the same arc $BD$ , and the angles $3$ and $4$ are equal as vertical. The triangles $ADE$ and $CBE$ are similar (according to the first triangle similarity criterion).

Then $\dfrac(AE)(EC) = \dfrac(DE)(BE)$, whence $AE\cdot BE = CE\cdot DE$ .

Tangent and secant theorem

The square of a tangent segment is equal to the product of the 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 general, $\angle BCA = 0.5\cdot\buildrel\smile\over(AB)$. According to the angle theorem between a tangent and a secant, $\angle BAM = 0.5\cdot\buildrel\smile\over(AB) = \angle BCA$. Thus, the triangles $MBA$ and $MCA$ are similar in 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 the secant drawn from the point $O$ and its outer part does not depend on the choice of the secant drawn from the point $O$ .

Let us recall the cases of mutual arrangement of a straight line and a circle.

Given a circle with center O and radius r. The line P, the distance from the center to the line, that is, the perpendicular OM, is equal to d.

Case 1- the distance from the center of the circle to the straight line is less than the radius of the circle:

We have proved that in the case when the distance d is less than the radius of the circle r, the line and the circle have only two common points (Fig. 1).

Rice. 1. Case 1 illustration

Case two- the distance from the center of the circle to the straight line is equal to the radius of the circle:

We have proved that in this case the common point is unique (Fig. 2).

Rice. 2. Case 2 illustration

Case 3- the distance from the center of the circle to the straight line is greater than the radius of the circle:

We have proved that in this case the circle and the line do not have common points (Fig. 3).

Rice. 3. Case 3 illustration

In this lesson, we are interested in the second case, when the line and the circle have a single common point.

Definition:

A line that has a single common point with a circle is called a tangent to the circle, a common point is called a point of contact between the line and the circle.

The straight line p is a tangent, the point A is the point of contact (Fig. 4).

Rice. 4. Tangent

Theorem:

The tangent to the circle is perpendicular to the radius drawn to the point of contact (Fig. 5).

Rice. 5. Illustration for the theorem

Proof:

On the contrary, let OA not be perpendicular to the straight line p. In this case, let's drop the perpendicular from the point O to the line p, which will be the distance from the center of the circle to the line:

From a right triangle we can say that the hypotenuse OH is less than the leg OA, that is, the line and the circle have two common points, the line p is a secant. Thus, we have obtained a contradiction, which means that the theorem is proved.

Rice. 6. Illustration for the theorem

The converse theorem is also true.

Theorem:

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

Proof:

Since the line is perpendicular to the radius, the distance OA is the distance from the line to the center of the circle and it is equal to the radius: . That is, and in this case, as we previously argued, the line and the circle have the only common point - this is point A, so the line p is tangent to the circle by definition (Fig. 7).

Rice. 7. Illustration for the theorem

The direct and inverse theorems can be combined as follows (Fig. 8):

Given a circle with center O, straight line p, radius OA

Rice. 8. Illustration for the theorem

Theorem:

A line is tangent to a circle if and only if the radius drawn to the point of contact is perpendicular to it.

This theorem means that if the line is tangent, then the radius drawn to the point of contact is perpendicular to it, and vice versa, from the perpendicularity of OA and p it follows that p is tangent, that is, the line and the circle have a single common point.

Consider two tangents drawn from the same point to the circle.

Theorem:

The segments of tangents to a circle drawn from one point are equal and make equal angles with a straight line drawn through this point and the center of the circle.

Given a circle, center O, point A outside the circle. Two tangents are drawn from point A, points B and C are tangent points. It is required to prove that and that angles 3 and 4 are equal.

Rice. 9. Illustration for the theorem

Proof:

The proof is based on the equality of triangles . Explain the equality of triangles. They are rectangular, since the radius drawn to the point of contact is perpendicular to the tangent. Hence, the angles and are right and equal in . The legs OB and OS are equal, since they are the radius of the circle. Hypotenuse AO - common.

Thus, the triangles are equal in terms of the equality of the leg and hypotenuse. From this it is obvious that the legs AB and AC are also equal. Also, the angles opposite the equal sides are equal, which means that the angles and , are equal.

The theorem has been proven.

So, we got acquainted with the concept of a tangent to a circle, in the next lesson we will consider the degree measure of an arc of a circle.

Bibliography

Aleksandrov A.D. etc. Geometry Grade 8. - M.: Education, 2006.
Butuzov V.F., Kadomtsev S.B., Prasolov V.V. Geometry 8. - M.: Enlightenment, 2011.
Merzlyak A.G., Polonsky V.B., Yakir S.M. Geometry grade 8. - M.: VENTANA-GRAF, 2009.

Univer.omsk.su ().
Oldskola1.narod.ru ().
School6.aviel.ru ().

Homework

Atanasyan L.S., Butuzov V.F., Kadomtsev S.B. et al., Geometry 7-9, no. 634-637, p. 168.

points $x_0\in \mathbb(R)$ , and is differentiable in it: $f \in \mathcal(D)(x_0)$ . Tangent to the graph of a function $f$ at the point $x_0$ is called the graph of a linear function, given by the equation $y = f(x_0) + f"(x_0)(x-x_0),\quad x\in \mathbb(R)$ .

If the function

f

has at the point

x_0

infinite derivative

f"(x_0) = \pm\infty,

then the tangent line at this point is the vertical line given by the equation

x = x_0.

Comment

It follows directly from the definition that the graph of the tangent line passes through the point $(x_0,f(x_0))$ . Corner $\alpha$ between the tangent to the curve and the x-axis satisfies the equation

$\operatorname(tg)\,\alpha = f"(x_0)=k,$

where $\operatorname(tg)$ stands for tangent, and $\operatorname (k)$ - tangent slope coefficient. Derivative at a point $x_0$ is equal to angular coefficient tangent to the graph of the function $y = f(x)$ at this point.

Tangent as the limiting position of a secant

Let $f\colon U(x_0) \to \R$ and $x_1\in U(x_0).$ Then a straight line passing through the points $(x_0,f(x_0))$ and $(x_1,f(x_1))$ given by the equation

$y = f(x_0) + \frac(f(x_1) - f(x_0))(x_1 - x_0)(x-x_0).$

This line passes through the point $(x_0,f(x_0))$ for anyone $x_1\in U(x_0),$ and its angle of inclination $\alpha(x_1)$ satisfies the equation

$\operatorname(tg)\,\alpha(x_1) = \frac(f(x_1) - f(x_0))(x_1 - x_0).$

Due to the existence of the derivative of the function $f$ at the point $x_0,$ passing to the limit at $x_1\to x_0,$ we get that there is a limit

$\lim\limits_(x_1 \to x_0) \operatorname(tg)\,\alpha(x_1) = f"(x_0),$

and due to the continuity of the arc tangent and the limiting angle

$\alpha = \operatorname(arctg)\,f"(x_0).$

A line passing through a point $(x_0,f(x_0))$ and having a limit angle of inclination that satisfies $\operatorname(tg)\,\alpha = f"(x_0),$ is given by the tangent equation:

$y \u003d f (x_0) + f "(x_0) (x-x_0).$

Tangent to circle

A straight line that has one common point with a circle and lies in the same plane with it is called a tangent to the circle.

Properties

The tangent to the circle is perpendicular to the radius drawn to the point of contact.
The segments of tangents to the circle drawn from one point are equal and make equal angles with the line passing through this point and the center of the circle.
The length of the segment of the tangent drawn to a circle of unit radius, taken between the point of tangency and the point of intersection of the tangent with the ray drawn from the center of the circle, is the tangent of the angle between this ray and the direction from the center of the circle to the point of tangency. "Tangens" from lat. tangents- "tangent".

Variations and Generalizations

One-sided semi-tangents

If there is a right derivative $f"_+(x_0)< \infty,$ then right semitangent to the graph of the function $f$ at the point $x_0$ called beam

y = f(x_0) + f"_+(x_0)(x - x_0),\quad x \geqslant x_0.

If there is a left derivative $f"_-(x_0)< \infty,$ then left semitangent to the graph of the function $f$ at the point $x_0$ called beam

y = f(x_0) + f"_-(x_0)(x - x_0),\quad x \leqslant x_0.

If there is an infinite right derivative $f"_+(x_0) = +\infty\; (-\infty),$ $f$ at the point $x_0$ called beam

x = x_0, \;  y \geqslant f(x_0)\;  (y\leqslant f(x_0)).

If there is an infinite left derivative $f"_-(x_0) = +\infty\; (-\infty),$ then the right semitangent to the graph of the function $f$ at the point $x_0$ called beam

x = x_0, \;  y \leqslant f(x_0)\;  (y\geqslant f(x_0)).

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Literature

Toponogov V. A. Differential geometry of curves and surfaces. - Fizmatkniga, 2012. - ISBN 9785891552135.
// Encyclopedic Dictionary of Brockhaus and Efron: in 86 volumes (82 volumes and 4 additional). - St. Petersburg. , 1890-1907.

An excerpt characterizing the tangent line

- In places! - shouted a young officer at the soldiers gathered around Pierre. This young officer, apparently, performed his position for the first or second time, and therefore treated both the soldiers and the commander with particular distinctness and uniformity.
The erratic firing of cannons and rifles intensified throughout the field, especially to the left, where Bagration's flashes were, but because of the smoke of shots from the place where Pierre was, it was almost impossible to see anything. Moreover, observations of how, as it were, a family (separated from all others) circle of people who were on the battery, absorbed all the attention of Pierre. His first unconsciously joyful excitement, produced by the sight and sounds of the battlefield, was now replaced, especially after the sight of this lonely soldier lying in the meadow, by another feeling. Sitting now on the slope of the ditch, he watched the faces around him.
By ten o'clock, twenty people had already been carried away from the battery; two guns were broken, more and more shells hit the battery and flew, buzzing and whistling, long-range bullets. But the people who were on the battery did not seem to notice this; cheerful conversation and jokes were heard from all sides.
- Chinenko! - the soldier shouted at the approaching, whistling grenade. - Not here! To the infantry! - another added with a laugh, noticing that the grenade flew over and hit the ranks of the cover.
- What, friend? - laughed another soldier at the crouching peasant under the flying cannonball.
Several soldiers gathered at the rampart, looking at what was happening ahead.
“And they took off the chain, you see, they went back,” they said, pointing over the shaft.
“Look at your business,” the old non-commissioned officer shouted at them. - They went back, which means there is work back. - And the non-commissioned officer, taking one of the soldiers by the shoulder, pushed him with his knee. Laughter was heard.
- Roll on to the fifth gun! shouted from one side.
“Together, more amicably, in burlatski,” the cheerful cries of those who changed the gun were heard.
“Ay, I almost knocked off our master’s hat,” the red-faced joker laughed at Pierre, showing his teeth. “Oh, clumsy,” he added reproachfully to the ball that had fallen into the wheel and leg of a man.
- Well, you foxes! another laughed at the squirming militiamen who were entering the battery for the wounded.
- Al is not tasty porridge? Ah, crows, swayed! - they shouted at the militia, who hesitated in front of a soldier with a severed leg.
“Something like that, little one,” the peasants mimicked. - They don't like passion.
Pierre noticed how after each shot that hit, after each loss, a general revival flared up more and more.
As from an advancing thundercloud, more and more often, brighter and brighter flashed on the faces of all these people (as if in repulse to what was happening) lightning bolts of hidden, flaring fire.
Pierre did not look ahead on the battlefield and was not interested in knowing what was happening there: he was completely absorbed in contemplating this, more and more burning fire, which in the same way (he felt) flared up in his soul.
At ten o'clock the infantry soldiers, who were ahead of the battery in the bushes and along the Kamenka River, retreated. From the battery it was visible how they ran back past it, carrying the wounded on their guns. Some general with his retinue entered the mound and, after talking with the colonel, looking angrily at Pierre, went down again, ordering the infantry cover, which was standing behind the battery, to lie down so as to be less exposed to shots. Following this, in the ranks of the infantry, to the right of the battery, a drum was heard, shouts of command, and from the battery it was clear how the ranks of the infantry moved forward.
Pierre looked over the shaft. One face in particular caught his eye. It was an officer who, with a pale young face, was walking backwards, carrying a lowered sword, and looking around uneasily.
The ranks of infantry soldiers disappeared into the smoke, their long-drawn cry and frequent firing of guns were heard. A few minutes later, crowds of wounded and stretchers passed from there. Shells began to hit the battery even more often. Several people lay uncleaned. Near the cannons, the soldiers moved busier and more lively. No one paid any attention to Pierre anymore. Once or twice he was angrily shouted at for being on the road. The senior officer, with a frown on his face, moved with large, quick steps from one gun to another. The young officer, flushed even more, commanded the soldiers even more diligently. Soldiers fired, turned, loaded and did their job with intense panache. They bounced along the way, as if on springs.