How to prove that a linear angle is a dihedral angle. Math lesson notes "Dihedral angle"

$\blacktriangleright$ Dihedral angle is an angle formed by two half-planes and a straight line $a$, which is their common boundary.

$\blacktriangleright$ To find the angle between the planes $\xi$ and $\pi$ , you need to find the linear angle (and spicy or straight) dihedral angle formed by the planes $\xi$ and $\pi$ :

Step 1: let $\xi\cap\pi=a$ (the line of intersection of the planes). In the plane $\xi$ we mark an arbitrary point $F$ and draw $FA\perp a$ ;

Step 2: carry out $FG\perp \pi$ ;

Step 3: according to TTP ($FG$ – perpendicular, $FA$ – oblique, $AG$ – projection) we have: $AG\perp a$ ;

Step 4: The angle $\angle FAG$ is called the linear angle of the dihedral angle formed by the planes $\xi$ and $\pi$ .

Note that the triangle $AG$ is right-angled.
Note also that the plane $AFG$ constructed in this way is perpendicular to both planes $\xi$ and $\pi$ . Therefore, we can say it differently: angle between planes$\xi$ and $\pi$ is the angle between two intersecting lines $c\in \xi$ and $b\in\pi$ forming a plane perpendicular to and $\xi\ ) , and \(\pi$ .

Task 1 #2875

Task level: More difficult than the Unified State Exam

Given a quadrangular pyramid, all edges of which are equal, and the base is a square. Find $6\cos \alpha$ , where $\alpha$ is the angle between its adjacent side faces.

Let $SABCD$ be a given pyramid ($S$ is a vertex) whose edges are equal to $a$ . Consequently, all side faces are equal equilateral triangles. Let's find the angle between the faces $SAD$ and $SCD$ .

Let's do $CH\perp SD$ . Because $\triangle SAD=\triangle SCD$, then $AH$ will also be the height of $\triangle SAD$ . Therefore, by definition, $\angle AHC=\alpha$ is the linear angle of the dihedral angle between the faces $SAD$ and $SCD$ .
Since the base is a square, then $AC=a\sqrt2$ . Note also that $CH=AH$ is the height of an equilateral triangle with side $a$, therefore, $CH=AH=\frac(\sqrt3)2a$ .
Then, by the cosine theorem from $\triangle AHC$ : \[\cos \alpha=\dfrac(CH^2+AH^2-AC^2)(2CH\cdot AH)=-\dfrac13 \quad\Rightarrow\quad 6\cos\alpha=-2.\]

Answer: -2

Task 2 #2876

Task level: More difficult than the Unified State Exam

The planes $\pi_1$ and $\pi_2$ intersect at an angle whose cosine is equal to $0.2$. The planes $\pi_2$ and $\pi_3$ intersect at right angles, and the line of intersection of the planes $\pi_1$ and $\pi_2$ is parallel to the line of intersection of the planes $\pi_2$ and $\ pi_3$ . Find the sine of the angle between the planes $\pi_1$ and $\pi_3$ .

Let the line of intersection of $\pi_1$ and $\pi_2$ be a straight line $a$, the line of intersection of $\pi_2$ and $\pi_3$ be a straight line $b$, and the line of intersection $\pi_3$ and $\pi_1$ – straight line $c$ . Since $a\parallel b$ , then $c\parallel a\parallel b$ (according to the theorem from the section of the theoretical reference “Geometry in space” $\rightarrow$ “Introduction to stereometry, parallelism”).

Let's mark the points $A\in a, B\in b$ so that $AB\perp a, AB\perp b$ (this is possible since $a\parallel b$ ). Let us mark $C\in c$ so that $BC\perp c$ , therefore, $BC\perp b$ . Then $AC\perp c$ and $AC\perp a$ .
Indeed, since $AB\perp b, BC\perp b$ , then $b$ is perpendicular to the plane $ABC$ . Since $c\parallel a\parallel b$, then the lines $a$ and $c$ are also perpendicular to the plane $ABC$, and therefore to any line from this plane, in particular, the line \ (AC\) .

It follows that $\angle BAC=\angle (\pi_1, \pi_2)$, $\angle ABC=\angle (\pi_2, \pi_3)=90^\circ$, $\angle BCA=\angle (\pi_3, \pi_1)$. It turns out that $\triangle ABC$ is rectangular, which means \[\sin \angle BCA=\cos \angle BAC=0.2.\]

Answer: 0.2

Task 3 #2877

Task level: More difficult than the Unified State Exam

Given straight lines $a, b, c$ intersecting at one point, and the angle between any two of them is equal to $60^\circ$ . Find $\cos^(-1)\alpha$ , where $\alpha$ is the angle between the plane formed by lines $a$ and $c$ and the plane formed by lines $b\ ) and \(c$ . Give your answer in degrees.

Let the lines intersect at the point $O$ . Since the angle between any two of them is equal to $60^\circ$, then all three straight lines cannot lie in the same plane. Let us mark the point $A$ on the line $a$ and draw $AB\perp b$ and $AC\perp c$ . Then $\triangle AOB=\triangle AOC$ as rectangular along the hypotenuse and acute angle. Therefore, $OB=OC$ and $AB=AC$ .
Let's do $AH\perp (BOC)$ . Then by the theorem about three perpendiculars $HC\perp c$ , $HB\perp b$ . Since $AB=AC$ , then $\triangle AHB=\triangle AHC$ as rectangular along the hypotenuse and leg. Therefore, $HB=HC$ . This means that $OH$ is the bisector of the angle $BOC$ (since the point $H$ is equidistant from the sides of the angle).

Note that in this way we also constructed the linear angle of the dihedral angle formed by the plane formed by the lines $a$ and $c$ and the plane formed by the lines $b$ and $c$ . This is the angle $ACH$ .

Let's find this angle. Since we chose the point $A$ arbitrarily, let us choose it so that $OA=2$ . Then in rectangular $\triangle AOC$ : \[\sin 60^\circ=\dfrac(AC)(OA) \quad\Rightarrow\quad AC=\sqrt3 \quad\Rightarrow\quad OC=\sqrt(OA^2-AC^2)=1.\ ] Since $OH$ is a bisector, then $\angle HOC=30^\circ$ , therefore, in a rectangular $\triangle HOC$ : \[\mathrm(tg)\,30^\circ=\dfrac(HC)(OC)\quad\Rightarrow\quad HC=\dfrac1(\sqrt3).\] Then from the rectangular $\triangle ACH$ : \[\cos\angle \alpha=\cos\angle ACH=\dfrac(HC)(AC)=\dfrac13 \quad\Rightarrow\quad \cos^(-1)\alpha=3.\]

Answer: 3

Task 4 #2910

Task level: More difficult than the Unified State Exam

The planes $\pi_1$ and $\pi_2$ intersect along the straight line $l$ on which the points $M$ and $N$ lie. The segments $MA$ and $MB$ are perpendicular to the straight line $l$ and lie in the planes $\pi_1$ and $\pi_2$ respectively, and $MN = 15$ , $AN = 39$ , $BN = 17$ , $AB = 40$ . Find $3\cos\alpha$ , where $\alpha$ is the angle between the planes $\pi_1$ and $\pi_2$ .

The triangle $AMN$ is right-angled, $AN^2 = AM^2 + MN^2$, whence \ The triangle $BMN$ is right-angled, $BN^2 = BM^2 + MN^2$, from which \We write the cosine theorem for the triangle $AMB$: \ Then \ Since the angle $\alpha$ between the planes is sharp corner, and $\angle AMB$ turned out to be obtuse, then $\cos\alpha=\dfrac5(12)$ . Then \

Answer: 1.25

Task 5 #2911

Task level: More difficult than the Unified State Exam

$ABCDA_1B_1C_1D_1$ is a parallelepiped, $ABCD$ is a square with side $a$, point $M$ is the base of the perpendicular dropped from the point $A_1$ to the plane $(ABCD)$ , in addition, $M$ is the point of intersection of the diagonals of the square $ABCD$ . It is known that $A_1M = \dfrac(\sqrt(3))(2)a$. Find the angle between the planes $(ABCD)$ and $(AA_1B_1B)$ . Give your answer in degrees.

Let's construct $MN$ perpendicular to $AB$ as shown in the figure.

Since $ABCD$ is a square with side $a$ and $MN\perp AB$ and $BC\perp AB$ , then $MN\parallel BC$ . Since $M$ is the point of intersection of the diagonals of the square, then $M$ is the middle of $AC$, therefore, $MN$ is the middle line and $MN =\frac12BC= \frac(1)(2)a$.
$MN$ is the projection of $A_1N$ onto the plane $(ABCD)$, and $MN$ is perpendicular to $AB$, then, by the theorem of three perpendiculars, $A_1N$ is perpendicular to $AB $ and the angle between the planes $(ABCD)$ and $(AA_1B_1B)$ is $\angle A_1NM$ .
\[\mathrm(tg)\, \angle A_1NM = \dfrac(A_1M)(NM) = \dfrac(\frac(\sqrt(3))(2)a)(\frac(1)(2)a) = \sqrt(3)\qquad\Rightarrow\qquad\angle A_1NM = 60^(\circ)\]

Answer: 60

Task 6 #1854

Task level: More difficult than the Unified State Exam

In a square $ABCD$ : $O$ – the point of intersection of the diagonals; $S$ – does not lie in the plane of the square, $SO \perp ABC$ . Find the angle between the planes $ASD$ and $ABC$ if $SO = 5$ and $AB = 10$ .

Right triangles $\triangle SAO$ and $\triangle SDO$ are equal in two sides and the angle between them ($SO \perp ABC$ $\Rightarrow$ $\angle SOA = \angle SOD = 90^\circ$; $AO = DO$ , because $O$ – point of intersection of the diagonals of the square, $SO$ – common side) $\Rightarrow$ $AS = SD$ $\Rightarrow$ $\triangle ASD$ – isosceles. Point $K$ is the middle of $AD$, then $SK$ is the height in the triangle $\triangle ASD$, and $OK$ is the height in the triangle $AOD$ $\ Rightarrow$ plane $SOK$ is perpendicular to planes $ASD$ and $ABC$ $\Rightarrow$ $\angle SKO$ – linear angle equal to the desired dihedral angle.

In $\triangle SKO$ : $OK = \frac(1)(2)\cdot AB = \frac(1)(2)\cdot 10 = 5 = SO$$\Rightarrow$ $\triangle SOK$ – isosceles right triangle $\Rightarrow$ $\angle SKO = 45^\circ$ .

Answer: 45

Task 7 #1855

Task level: More difficult than the Unified State Exam

Right triangles $\triangle SAO$ , $\triangle SDO$ , $\triangle SOB$ and $\triangle SOC$ are equal in two sides and the angle between them ($SO \perp ABC$ $\Rightarrow$ $\angle SOA = \angle SOD = \angle SOB = \angle SOC = 90^\circ$; $AO = OD = OB = OC$, because $O$ – point of intersection of the diagonals of the square, $SO$ – common side) $\Rightarrow$ $AS = DS = BS = CS$ $\Rightarrow$ $\triangle ASD$ and $\triangle BSC$ are isosceles. Point $K$ is the middle of $AD$, then $SK$ is the height in the triangle $\triangle ASD$, and $OK$ is the height in the triangle $AOD$ $\ Rightarrow$ plane $SOK$ is perpendicular to plane $ASD$ . Point $L$ is the middle of $BC$, then $SL$ is the height in the triangle $\triangle BSC$, and $OL$ is the height in the triangle $BOC$ $\ Rightarrow$ plane $SOL$ (aka plane $SOK$) is perpendicular to the plane $BSC$ . Thus, we obtain that $\angle KSL$ is a linear angle equal to the desired dihedral angle.

$KL = KO + OL = 2\cdot OL = AB = 10$$\Rightarrow$ $OL = 5$ ; $SK = SL$ – equal heights isosceles triangles, which can be found using the Pythagorean theorem: $SL^2 = SO^2 + OL^2 = 5^2 + 5^2 = 50$. It can be noticed that $SK^2 + SL^2 = 50 + 50 = 100 = KL^2$$\Rightarrow$ for a triangle $\triangle KSL$ the inverse Pythagorean theorem holds $\Rightarrow$ $\triangle KSL$ – right triangle $\Rightarrow$ $\angle KSL = 90^\ circ$ .

Answer: 90

Preparing students to take the Unified State Exam in mathematics, as a rule, begins with repeating basic formulas, including those that allow you to determine the angle between planes. Despite the fact that this section of geometry is covered in sufficient detail within school curriculum, many graduates need to repeat basic material. Understanding how to find the angle between planes, high school students will be able to quickly calculate the correct answer when solving a problem and count on receiving decent scores on the results of passing the unified state exam.

Main nuances

To ensure that the question of how to find a dihedral angle does not cause difficulties, we recommend following a solution algorithm that will help you cope with Unified State Examination tasks.

First you need to determine the straight line along which the planes intersect.

Then you need to select a point on this line and draw two perpendiculars to it.

Next step- finding trigonometric function dihedral angle formed by perpendiculars. The most convenient way to do this is with the help of the resulting triangle, of which the angle is a part.

The answer will be the value of the angle or its trigonometric function.

Preparing for the exam test with Shkolkovo is the key to your success

During classes the day before passing the Unified State Exam Many schoolchildren are faced with the problem of finding definitions and formulas that allow them to calculate the angle between 2 planes. A school textbook is not always at hand exactly when needed. And in order to find the necessary formulas and examples of their correct application, including for finding the angle between planes on the Internet online, sometimes you need to spend a lot of time.

The Shkolkovo mathematical portal offers a new approach to preparing for the state exam. Classes on our website will help students identify the most difficult sections for themselves and fill gaps in knowledge.

We have prepared and clearly presented everything required material. Basic definitions and formulas are presented in the “Theoretical Information” section.

In order to better understand the material, we also suggest practicing the appropriate exercises. A large selection of tasks of varying degrees of complexity, for example, on, is presented in the “Catalogue” section. All tasks contain a detailed algorithm for finding the correct answer. The list of exercises on the website is constantly supplemented and updated.

While practicing solving problems that require finding the angle between two planes, students have the opportunity to save any task online as “Favorites.” Thanks to this, they will be able to return to it as many times as necessary and discuss the progress of its solution with school teacher or a tutor.

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Concept of dihedral angle

To introduce the concept of a dihedral angle, let us first recall one of the axioms of stereometry.

Any plane can be divided into two half-planes of the line $a$ lying in this plane. In this case, points lying in the same half-plane are located on one side of the straight line $a$, and points lying in different half-planes are on the same side. different sides from straight line $a$ (Fig. 1).

Picture 1.

The principle of constructing a dihedral angle is based on this axiom.

Definition 1

The figure is called dihedral angle, if it consists of a line and two half-planes of this line that do not belong to the same plane.

In this case, the half-planes of the dihedral angle are called edges, and the straight line separating the half-planes is dihedral edge(Fig. 1).

Figure 2. Dihedral angle

Degree measure of dihedral angle

Definition 2

Let us choose an arbitrary point $A$ on the edge. The angle between two straight lines lying in different half-planes, perpendicular to an edge and intersecting at point $A$ is called linear dihedral angle(Fig. 3).

Figure 3.

Obviously, every dihedral angle has an infinite number of linear angles.

Theorem 1

All linear angles of one dihedral angle are equal to each other.

Proof.

Let's consider two linear angles $AOB$ and $A_1(OB)_1$ (Fig. 4).

Figure 4.

Since the rays $OA$ and $(OA)_1$ lie in the same half-plane $\alpha $ and are perpendicular to the same straight line, then they are codirectional. Since the rays $OB$ and $(OB)_1$ lie in the same half-plane $\beta $ and are perpendicular to the same straight line, then they are codirectional. Hence

\[\angle AOB=\angle A_1(OB)_1\]

Due to the arbitrariness of the choice of linear angles. All linear angles of one dihedral angle are equal to each other.

The theorem has been proven.

Definition 3

The degree measure of a dihedral angle is the degree measure of the linear angle of a dihedral angle.

Sample problems

Example 1

Let us be given two non-perpendicular planes $\alpha $ and $\beta $ that intersect along the straight line $m$. Point $A$ belongs to the plane $\beta$. $AB$ is perpendicular to line $m$. $AC$ is perpendicular to the plane $\alpha $ (point $C$ belongs to $\alpha $). Prove that angle $ABC$ is a linear angle of a dihedral angle.

Proof.

Let's draw a picture according to the conditions of the problem (Fig. 5).

Figure 5.

To prove it, recall the following theorem

Theorem 2: A straight line passing through the base of an inclined one is perpendicular to it, perpendicular to its projection.

Since $AC$ is perpendicular to the plane $\alpha $, then point $C$ is the projection of point $A$ onto the plane $\alpha $. Therefore, $BC$ is a projection of the oblique $AB$. By Theorem 2, $BC$ is perpendicular to the edge of the dihedral angle.

Then, angle $ABC$ satisfies all the requirements for defining a linear dihedral angle.

Example 2

The dihedral angle is $30^\circ$. On one of the faces lies a point $A$, which is located at a distance of $4$ cm from the other face. Find the distance from the point $A$ to the edge of the dihedral angle.

Solution.

Let's look at Figure 5.

By condition, we have $AC=4\cm$.

By the definition of the degree measure of a dihedral angle, we have that the angle $ABC$ is equal to $30^\circ$.

Triangle $ABC$ is a right triangle. By definition of the sine of an acute angle

\[\frac(AC)(AB)=sin(30)^0\] \[\frac(5)(AB)=\frac(1)(2)\] \

Lesson topic: “Dihedral angle.”

The purpose of the lesson: introduction of the concept of dihedral angle and its linear angle.

Tasks:

Educational: consider tasks on the application of these concepts, develop the constructive skill of finding the angle between planes;

Developmental: development creative thinking students, personal self-development of students, development of students’ speech;

Educational: nurturing a culture of mental work, communicative culture, reflective culture.

Lesson type: lesson in learning new knowledge

Teaching methods: explanatory and illustrative

Equipment: computer, interactive whiteboard.

Literature:

Geometry. Grades 10-11: textbook. for 10-11 grades. general education institutions: basic and profile. levels / [L. S. Atanasyan, V. F. Butuzov, S. B. Kadomtsev, etc.] - 18th ed. – M.: Education, 2009. – 255 p.

Lesson plan:

Organizing time(2 minutes)

Updating knowledge (5 min)

Learning new material (12 min)

Reinforcement of learned material (21 min)

Homework (2 min)

Summing up (3 min)

During the classes:

1. Organizational moment.

Includes the teacher greeting the class, preparing the room for the lesson, and checking on absentees.

2. Updating basic knowledge.

Teacher: In the last lesson you wrote independent work. In general, the work was written well. Now let's repeat it a little. What is an angle in a plane called?

Student: An angle on a plane is a figure formed by two rays emanating from one point.

Teacher: What is the angle between lines in space called?

Student: The angle between two intersecting lines in space is the smallest of the angles formed by the rays of these lines with the vertex at the point of their intersection.

Student: The angle between intersecting lines is the angle between intersecting lines, respectively, parallel to the data.

Teacher: What is the angle between a straight line and a plane called?

Student: The angle between a straight line and a planeAny angle between a straight line and its projection onto this plane is called.

3. Studying new material.

Teacher: In stereometry, along with such angles, another type of angle is considered - dihedral angles. You probably already guessed what the topic of today's lesson is, so open your notebooks, write down today's date and the topic of the lesson.

Write on the board and in notebooks:

10.12.14.

Dihedral angle.

Teacher : To introduce the concept of a dihedral angle, it should be recalled that any straight line drawn in a given plane divides this plane into two half-planes(Fig. 1, a)

Teacher : Let’s imagine that we have bent the plane along a straight line so that two half-planes with a boundary no longer lie in the same plane (Fig. 1, b). The resulting figure is the dihedral angle. A dihedral angle is a figure formed by a straight line and two half-planes with a common boundary that do not belong to the same plane. The half-planes forming a dihedral angle are called its faces. A dihedral angle has two sides, hence the name dihedral angle. The straight line - the common boundary of the half-planes - is called the edge of the dihedral angle. Write the definition in your notebook.

A dihedral angle is a figure formed by a straight line and two half-planes with a common boundary that do not belong to the same plane.

Teacher : In everyday life, we often encounter objects that have the shape of a dihedral angle. Give examples.

Student : Half-opened folder.

Student : The wall of the room is together with the floor.

Student : Gable roofs of buildings.

Teacher : Right. And there are a huge number of such examples.

Teacher : As you know, angles in a plane are measured in degrees. You probably have a question, how are dihedral angles measured? This is done as follows.Let's mark some point on the edge of the dihedral angle and draw a ray perpendicular to the edge from this point on each face. The angle formed by these rays is called the linear angle of the dihedral angle. Make a drawing in your notebooks.

Write on the board and in notebooks.

ABOUT ∈ a, JSC ⊥ a, VO ⊥ a, SABD– dihedral angle,∠ AOB– linear angle of the dihedral angle.

Teacher : All linear angles of a dihedral angle are equal. Make yourself another drawing like this.

Teacher : Let's prove it. Consider two linear angles AOB andPQR. Rays OA andQPlie on the same face and are perpendicularOQ, which means they are co-directed. Similarly, the rays OB andQRco-directed. Means,∠ AOB= ∠ PQR(like angles with aligned sides).

Teacher : Well, now the answer to our question is how the dihedral angle is measured.The degree measure of a dihedral angle is the degree measure of its linear angle. Redraw the images of an acute, right and obtuse dihedral angle from the textbook on page 48.

4. Consolidation of the studied material.

Teacher : Make drawings for the tasks.

№ 1 . Given: ΔABC, AC = BC, AB lies in the planeα, CD ⊥ α, C∉ α. Construct linear angle of dihedral angleCABD.

Student : Solution:C.M. ⊥ AB, DC ⊥ AB.∠ CMD - sought after.

№ 2. Given: ΔABC, ∠ C= 90°, BC lies on the planeα, JSC⊥ α, A∈ α.

Construct linear angle of dihedral angleABCO.

Student : Solution:AB ⊥ B.C., JSC⊥ BC means OS⊥ Sun.∠ ACO - sought after.

№ 3 . Given: ΔABC, ∠ C = 90°, AB lies in the planeα, CD⊥ α, C∉ α. Buildlinear dihedral angleDABC.

Student : Solution: CK ⊥ AB, DC ⊥ AB,DK ⊥ AB means∠ DKC - sought after.

№ 4 . Given:DABC- tetrahedron,DO⊥ ABC.Construct the linear angle of the dihedral angleABCD.

Student : Solution:DM ⊥ Sun,DO ⊥ VS means OM⊥ Sun;∠ OMD - sought after.

5. Summing up.

Teacher: What new did you learn in class today?

Students : What is called dihedral angle, linear angle, how is dihedral angle measured.

Teacher : What did they repeat?

Students : What is called an angle on a plane; angle between straight lines.

6.Homework.

Write on the board and in your diaries: paragraph 22, No. 167, No. 170.

Dihedral angle. Linear dihedral angle. A dihedral angle is a figure formed by two half-planes that do not belong to the same plane and have a common boundary - straight line a. The half-planes that form a dihedral angle are called its faces, and the common boundary of these half-planes is called an edge of the dihedral angle. The linear angle of a dihedral angle is an angle whose sides are the rays along which the faces of the dihedral angle are intersected by a plane perpendicular to the edge of the dihedral angle. Each dihedral angle has any number of linear angles: through each point of an edge one can draw a plane perpendicular to this edge; The rays along which this plane intersects the faces of a dihedral angle form linear angles.

All linear angles of a dihedral angle are equal to each other. Let us prove that if the dihedral angles formed by the plane of the base of the pyramid CABC and the planes of its lateral faces are equal, then the base of the perpendicular drawn from vertex K is the center of the inscribed circle in triangle ABC.

Proof. First of all, let's construct linear angles of equal dihedral angles. By definition, the plane of a linear angle must be perpendicular to the edge of the dihedral angle. Therefore, the edge of a dihedral angle must be perpendicular to the sides of the linear angle. If KO is perpendicular to the base plane, then we can draw OR perpendicular AC, OR perpendicular SV, OQ perpendicular AB, and then connect points P, Q, R WITH point K. Thus, we will construct a projection of inclined RK, QK, RK so that the edges AC, NE, AB are perpendicular to these projections. Consequently, these edges are perpendicular to the inclined ones themselves. And therefore the planes of triangles ROK, QOK, ROK are perpendicular to the corresponding edges of the dihedral angle and form those equal linear angles that are mentioned in the condition. Right triangles ROK, QOK, ROK are congruent (since they have a common leg OK and the angles opposite to this leg are equal). Therefore, OR = OR = OQ. If we draw a circle with center O and radius OP, then the sides of triangle ABC are perpendicular to the radii OP, OR and OQ and therefore are tangent to this circle.

Perpendicularity of planes. The alpha and beta planes are called perpendicular if the linear angle of one of the dihedral angles formed at their intersection is equal to 90." Signs of perpendicularity of two planes If one of the two planes passes through a line perpendicular to the other plane, then these planes are perpendicular.

The figure shows a rectangular parallelepiped. Its bases are rectangles ABCD and A1B1C1D1. And the side ribs AA1 BB1, CC1, DD1 are perpendicular to the bases. It follows that AA1 is perpendicular to AB, i.e. the side face is a rectangle. Thus, it is possible to justify the properties rectangular parallelepiped: In a rectangular parallelepiped, all six faces are rectangles. In a rectangular parallelepiped, all six faces are rectangles. All dihedral angles of a rectangular parallelepiped are right angles. All dihedral angles of a rectangular parallelepiped are right angles.

Theorem Square of the diagonal of a rectangular parallelepiped equal to the sum squares of its three dimensions. Let us turn again to the figure, and prove that AC12 = AB2 + AD2 + AA12 Since edge CC1 is perpendicular to the base ABCD, angle ACC1 is right. From right triangle ACC1 using the Pythagorean theorem we obtain AC12=AC2+CC12. But AC is a diagonal of rectangle ABCD, so AC2 = AB2 + AD2. In addition, CC1 = AA1. Therefore AC12= AB2+AD2+AA12 The theorem is proven.