Properties of inclined lines emerging from one point. 1. A perpendicular is always shorter than an inclined one if they are drawn from the same point. 2. If the inclined ones are equal, then their projections are equal, and vice versa. 3. A larger oblique corresponds to a larger projection and vice versa.
Slide 10 from the presentation "Perpendicular and inclined to the plane". The size of the archive with the presentation is 327 KB.Geometry 10th grade
summary other presentations“Parallelogram problems” - Geometry. Dots. Height of the parallelogram. Square. Proof. Tangent to a circle. Signs of a parallelogram. Perimeter of a parallelogram. Circle. Part. Middle line. Centers of circles. Angles. Parallelogram. Find the area of the parallelogram. Two circles. Properties of a parallelogram. Sharp corner. Area of a parallelogram. Diagonals of a parallelogram. Diagonal. Quadrangle. Triangles.
“Methods for constructing sections” - Formation of skills in constructing sections. Let's consider four cases of constructing sections of a parallelepiped. Construct sections of the tetrahedron. Method interior design. Working with disks. The parallelepiped has six faces. Cutting plane. Construction of sections of polyhedra. The trace is the straight line of intersection of the section plane and the plane of any face of the polyhedron. Trace method. Memo.
““Regular polyhedra” 10th grade” - Predicted result. A tetrahedron described near the orbital sphere of Mars. Center O, axis a and plane. Faces of a polyhedron. Radiolaria. Content. Regular polyhedra. Regular polyhedra in Plato's philosophical picture of the world. Feodaria. Regular polyhedra are found in living nature. During the classes. A point (straight line, plane) is called a center (axis, plane). Which of the following geometric bodies is not a regular polyhedron.
“Determination of dihedral angles” - Point K is removed from each side. Points M and K lie in different faces. Degree measure corner. Property trihedral angle. Notes on problem solving. On one of the edges dihedral angle, equal to 30, point M is located. Construction linear angle. Draw a perpendicular. A straight line drawn in a given plane. Dihedral angles in pyramids. Problem solving. Point K. This pyramid. The point on the edge can be arbitrary.
“Methods for constructing sections of polyhedra” - Any plane. Artists. Laws of geometry. Blitz survey. Mutual arrangement plane and polyhedron. Construct a section of a polyhedron. Polygons. Axiomatic method. Tasks. Ship. Task. Axioms. Construction of sections of polyhedra. Sections by different planes. Ancient Chinese proverb. Independent work. Diagonal sections. Consolidation of acquired knowledge. Cutting plane.
“Equilateral Polygons” - Hexahedron (Cube) The cube is made up of six squares. Octahedron The octahedron is made up of eight equilateral triangles. A tetrahedron has 4 faces, 4 vertices and 6 edges. There are 5 types regular polyhedra. Regular Polygons. The dodecahedron has 12 faces, 20 vertices and 30 edges. The icosahedron has 20 faces, 12 vertices and 30 edges. Thus, a cube has 6 faces, 8 vertices and 12 edges. Tetrahedron The tetrahedron is made up of four equilateral triangles.
If through some point taken outside a line we draw a line perpendicular to it, then for brevity the segment from this point to the line is called in one word perpendicular.
Segment CO is perpendicular to line AB. Point O is called base of the perpendicular CO (rice).
If a straight line drawn through this point, intersects another line, but is not perpendicular to it, then its segment from a given point to the point of intersection with another line is called inclined to this line.
Segment BC - inclined to straight line AO. Point C is called basis inclined (fig.).
If we drop perpendiculars from the ends of some segment onto an arbitrary line, then the line segment enclosed between the bases of the perpendiculars is called projection of the segment to this straight line.
Segment АВ - projection of segment AB onto EC. The segment OM is also called the projection of the segment OM onto the EC.
The projection of the segment KP perpendicular to EC will be the point K (Fig.).
2. Properties of perpendicular and oblique.
Theorem 1. A perpendicular drawn from a point to a straight line is less than any oblique drawn from the same point to this straight line.
The segment AC (Fig.) is perpendicular to straight line OB, and AM is one of the inclined lines drawn from point A to straight line OB. It is required to prove that AM > AC.
In ΔMAC, the segment AM is the hypotenuse, and the hypotenuse is larger than each of the legs of this triangle. Therefore, AM > AC. Since we took the inclined AM arbitrarily, we can say that any inclined line to a straight line is greater than a perpendicular to this line (and a perpendicular is shorter than any inclined line) if they are drawn to it from the same point.
The converse statement is also true, namely: if the segment AC (Fig.) is less than any other segment connecting the point AC with any point on the straight line OB, then it is perpendicular to OB. In fact, segment AC cannot be inclined to OB, since then it would not be the shortest of the segments connecting point A with points of straight line OB. This means that it can only be perpendicular to OB.
The length of the perpendicular dropped from a given point to a straight line is taken as the distance from a given point to this straight line.
Theorem 2. If two oblique lines drawn to a line from the same point are equal, then their projections are equal.
Let BA and BC be inclined lines drawn from point B to straight line AC (Fig.), and AB = BC. It is necessary to prove that their projections are also equal.
To prove this, let us lower the perpendicular BO from point B to AC. Then AO and OS will be projections of inclined AB and BC onto straight line AC. Triangle ABC isosceles according to the theorem. VO is the height of this triangle. But the height is isosceles triangle, drawn to the base, is at the same time the median of this triangle.
Therefore AO = OS.
Theorem 3 (converse). If two oblique lines drawn to a straight line from the same point have equal projections, then they are equal to each other.
Let AC and CB be inclined to straight line AB (Fig.). CO ⊥ AB and AO = OB.
It is required to prove that AC = BC.
In right triangles AOC and BOC, the legs AO and OB are equal. CO is the common leg of these triangles. Therefore, ΔAOC = ΔBOC. From the equality of triangles it follows that AC = BC.
Theorem 4. If two inclined lines are drawn from the same point to a straight line, then the one that has a larger projection onto this straight line is larger.
Let AB and BC be inclined to straight line AO; VO ⊥ AO and AO>CO. It is required to prove that AB > BC.
1) Inclined ones are located on one side of the perpendicular.
Angle ACE external to right triangle SOV (fig.), and therefore ∠ASV > ∠SOV, i.e. he is stupid. It follows that AB > CB.
2) Inclined ones are located on both sides of the perpendicular. To prove this, let’s plot the segment OK = OS on AO from point O and connect point K to point B (Fig.). Then, by Theorem 3, we have: VC = BC, but AB > VC, therefore, AB > BC, i.e. the theorem is valid in this case as well.
Theorem 5 (converse). If two inclined lines are drawn from the same point to a straight line, then the larger inclined line also has a larger projection onto this straight line.
Let KS and BC be inclined to the straight line CV (Fig.), SO ⊥ CV and KS > BC. It is required to prove that KO > OB.
Between the segments KO and OB there can be only one of three relationships:
1) KO< ОВ,
2) KO = OV,
3) KO > OV.
KO cannot be less than OB, since then, according to Theorem 4, the inclined KS would be less than the inclined BC, and this contradicts the conditions of the theorem.
In the same way, KO cannot equal OB, since in this case, according to Theorem 3, KS = BC, which also contradicts the conditions of the theorem.
Consequently, only the last relation remains true, namely, that KO > OB.
A perpendicular dropped from a given point to given plane, is called a segment connecting a given point with a point in the plane and lying on a straight line, perpendicular to the plane. The end of this segment lying in the plane is called the base of the perpendicular. The distance from a point to a plane is the length of the perpendicular drawn from this point to the plane.
The slope drawn from a given point to a given plane is any segment that connects a given point to a point on the plane and is not perpendicular to this plane. The end of a segment lying in a plane is called the inclined base. A segment connecting the bases of a perpendicular and an oblique drawn from the same point is called an oblique projection.
In Figure 136, from point A, a perpendicular AB and an inclined AC are drawn to the plane. Point B is the base of the perpendicular, point C is the base of the inclined one, BC is the projection of the inclined AC onto plane a.
Since the distances from the points of a line to a plane parallel to it are the same, the distance from a line to a plane parallel to it is the distance from any point of it to this plane.
A straight line drawn on a plane through the base of an inclined plane perpendicular to its projection is also perpendicular to the inclined itself. And vice versa: if a straight line in a plane is perpendicular to an inclined one, then it is also perpendicular to the projection of the inclined one (the theorem of three perpendiculars).
In Figure 137, a perpendicular AB and an inclined AC are drawn to plane a. The straight line o, lying in the plane a, is perpendicular to BC - the projection of the inclined AC onto the plane a. According to T. 2.12, straight line a is perpendicular to inclined AC. If it were known that straight line a is perpendicular to inclined AC, then according to T. 2.12 it would be perpendicular to its projection - BC.
Example. Rectangular legs triangle ABC are equal to 16 and From the top right angle C is drawn perpendicular to the plane of this triangle CD = 35 m (Fig. 138). Find the distance from point D to the hypotenuse AB.
Solution. Let's do it. According to the condition, DC is perpendicular to the plane, i.e. DE is inclined, CE is its projection, therefore, by the theorem about three perpendiculars, it follows from the condition that
From we find To find the height CE in we find
On the other hand, where
From the Pythagorean theorem
46. Perpendicularity of planes.
Two intersecting planes are called perpendicular if any plane perpendicular to the line of intersection of these planes intersects them along perpendicular lines.
Figure 139 shows two planes that intersect along a straight line a. The plane y is perpendicular to the line a and intersects. In this case, the plane y intersects the plane a along the straight line c, and the plane intersects along the straight line d, and i.e., by definition
T. 2.13. If a plane passes through a line perpendicular to another plane, then these planes are perpendicular (a sign of perpendicularity of planes).
In Figure 140, the plane passes through a straight line, i.e., the plane is perpendicular.
TRIANGLES.
§ 31.PERPENDICULAR AND INCLINED TO A STRAIGHT.
1. Projection of a segment onto a straight line.
If through some point taken outside a line we draw a line perpendicular to it, then for brevity the segment from this point to the line is called in one word perpendicular.
Segment CO is perpendicular to line AB. Point O is called base of the perpendicular CO (drawing 168).
If a line drawn through a given point intersects another line, but is not perpendicular to it, then its segment from a given point to the point of intersection with another line is called inclined to this line.
Segment BC - inclined to straight line AO. Point C is called basis inclined (Fig. 169).
If we drop perpendiculars from the ends of some segment onto an arbitrary line, then the line segment enclosed between the bases of the perpendiculars is called projection of the segment to this straight line.
Segment A "B" is the projection of segment AB onto EC. Segment OM" is also called the projection of segment OM onto EC.
The projection of the segment KR perpendicular to the EU will be point K" (Fig. 170).
2. Properties of perpendicular and oblique.
Theorem 1. A perpendicular drawn from a point to a straight line is less than any oblique drawn from the same point to this straight line.
Segment AC (Fig. 171) is perpendicular to straight line OB, and AM is one of the inclined lines drawn from point A to straight line OB. It is required to prove that AM > AC.
IN /\ The MAC segment AM is the hypotenuse, and the hypotenuse is greater than each of the legs of this triangle (§ 30). Therefore, AM > AC. Since we took the inclined AM arbitrarily, we can say that any inclined line to a straight line is greater than a perpendicular to this line (and a perpendicular is shorter than any inclined line) if they are drawn to it from the same point.
The converse statement is also true, namely: if the segment AC (Fig. 171) is less than any other segment connecting the point AC with any point on the straight line OB, then it is perpendicular to OB. In fact, segment AC cannot be inclined to OB, since then it would not be the shortest of the segments connecting point A with points of straight line OB. This means that it can only be perpendicular to OB.
The length of the perpendicular dropped from a given point to a straight line is taken as the distance from a given point to this straight line.
Theorem 2. If two oblique lines drawn to a line from the same point are equal, then their projections are equal.
Let BA and BC be inclined lines drawn from point B to straight line AC (Fig. 172), and AB = BC. It is necessary to prove that their projections are also equal.
To prove this, let us lower the perpendicular BO from point B to AC. Then AO and OS will be projections of inclined AB and BC onto straight line AC. Triangle ABC is isosceles according to the theorem. VO is the height of this triangle. But the altitude in an isosceles triangle drawn to the base is at the same time the median of this triangle (§ 18).
Therefore AO = OS.
Theorem 3(reverse). If two oblique lines drawn to a straight line from the same point have equal projections, then they are equal to each other.
Let AC and CB be inclined to straight line AB (Fig. 173). CO_|_ AB and AO = OB.
It is required to prove that AC = BC.
In right triangles AOC and BOC, the legs AO and OB are equal. CO is the common leg of these triangles. Hence, /\ AOC = /\ SUN. From the equality of triangles it follows that AC = BC.
Theorem 4. If two inclined lines are drawn from the same point to a straight line, then the one that has a larger projection onto this straight line is larger.
Let AB and BC be inclined to straight line AO; VO_|_AO and AO>SO. It is required to prove that AB > BC.
1) Inclined ones are located on one side of the perpendicular.
Angle ACE is external with respect to the right triangle COB (Fig. 174), and therefore / DIA > / OWL, i.e. he is stupid. It follows that AB > CB.
2) Inclined ones are located on both sides of the perpendicular. To prove this, let us plot the segment OK = OS on AO from point O and connect point K with point B (Fig. 175). Then, by Theorem 3, we have: VC = BC, but AB > VC, therefore, AB > BC, i.e. the theorem is valid in this case as well.
Theorem 5(reverse). If two inclined lines are drawn from the same point to a straight line, then the larger inclined line also has a larger projection onto this straight line.
Let KS and BC be inclined to the straight line KB (Fig. 176), SO_|_KB and KS > BC. It is required to prove that KO > OB.
Between the segments KO and OB there can be only one of three relationships:
1) KO< ОВ,
2) KO = OV,
3) KO > OV.
KO cannot be less than OB, since then, according to Theorem 4, the inclined KS would be less than the inclined BC, and this contradicts the conditions of the theorem.
In the same way, KO cannot equal OB, since in this case, according to Theorem 3, KS = BC, which also contradicts the conditions of the theorem.
Consequently, only the last relation remains true, namely, that
KO > OV.
Theorem . If a perpendicular and inclined lines are drawn from one point outside the plane, then:
1) oblique ones having equal projections are equal;
2) of the two inclined ones, the one whose projection is larger is greater;
3) equal obliques have equal projections;
4) of the two projections, the one that corresponds to the larger oblique one is larger.
Three Perpendicular Theorem . In order for a straight line lying in a plane to be perpendicular to an inclined one, it is necessary and sufficient that this straight line be perpendicular to the projection of the inclined one (Fig. 12.3).
Theorem on the area of the orthogonal projection of a polygon onto a plane. The area of the orthogonal projection of a polygon onto a plane is equal to the product of the area of the polygon and the cosine of the angle between the plane of the polygon and the projection plane.
Example 1. Through a given point draw a straight line parallel to the given plane.
Solution. Analysis. Let's assume that the straight line is constructed (Fig. 12.4). A line is parallel to a plane if it is parallel to some line lying in the plane (based on the parallelism of the line and the plane). Two parallel lines lie in the same plane. This means that by constructing a plane passing through a given point and an arbitrary line in a given plane, it will be possible to construct a parallel line.
Construction.
1. On a plane we conduct a direct A.
3. In plane through the point A let's make a direct b, parallel to the line A.
4. A straight line has been built b, parallel to the plane .
Proof. Based on the parallelism of a straight line and a plane, a straight line b parallel to the plane , since it is parallel to the line A, belonging to the plane .
Study. The problem has an infinite number of solutions, since the straight line A in the plane is chosen randomly.
Example 2.
Determine at what distance from the plane the point is located A, if straight AB intersects the plane at an angle of 45º, the distance from the point A to the point IN, belonging to the plane, is equal to 
cm.
Solution. Let's make a drawing (Fig. 12.5):
AC– perpendicular to the plane
,
AB– inclined, angle ABC– angle between straight line AB and plane
. Triangle ABC– rectangular, 
because AC– perpendicular. The required distance from the point A to the plane - this is the leg AC right triangle. Knowing the angle 
and hypotenuse 
let's find the leg AC:
In response we get : AC = 3 cm.
Example 3. Determine at what distance from the plane of an isosceles triangle there is a point 13 cm distant from each of the vertices of the triangle if the base and height of the triangle are equal to 8 cm.
Solution. Let's make a drawing (Fig. 12.6). Dot S away from the points A, IN And WITH at the same distance. So, inclined S.A., S.B. And S.C. equal, SO– the common perpendicular of these inclined ones. By the theorem of obliques and projections AO = VO = CO.
Dot ABOUT– the center of a circle circumscribed about a triangle ABC. Let's find its radius:
Where Sun– base; AD– the height of a given isosceles triangle.
Finding the sides of a triangle ABC from a right triangle ABD according to the Pythagorean theorem:
Now we find OB:
Consider a triangle SOB:
S.B.= 13 cm, OB= 5 cm. Find the length of the perpendicular SO according to the Pythagorean theorem:
In response we get: SO= 12 cm.
Example 4. Given parallel planes And . Through the point M, which does not belong to any of them, straight lines are drawn A And b, which intersect the plane at points A 1 and IN 1 and the plane – at points A 2 and IN 2. Find A 1 IN 1 if it is known that MA 1 = 8 cm, A 1 A 2 = 12 cm, A 2 IN 2 = 25 cm.
Solution. Since the condition does not say how the point is located relative to both planes M, then two options are possible: (Fig. 12.7, a, b). Let's look at each of them. Two intersecting lines A And b define a plane. This plane intersects two parallel planes And along parallel lines A 1 IN 1 and A 2 IN 2 according to Theorem 5 about parallel lines and parallel planes.
Triangles MA 1 IN 1 and MA 2 IN 2 are similar (angles A 2 MV 2 and A 1 MV 1 – vertical, corners MA 1 IN 1 and MA 2 IN 2 – internal crosswise lying with parallel lines A 1 IN 1 and A 2 IN 2 and secant A 1 A 2). From the similarity of triangles follows the proportionality of the sides:
From here 
Option a):
Option b):
We get the answer: 10 cm and 50 cm.
Example 5. Through the point A plane a direct line was drawn AB, forming an angle with the plane . Via direct AB a plane is drawn , forming with a plane corner . Find the angle between the projection of a straight line AB to the plane and plane .
Solution.
Let's make a drawing (Fig. 12.8). From point IN drop the perpendicular to the plane
.
Linear dihedral angle between planes
And
- this is the angle 
Straight ADDBC, based on the perpendicularity of a straight line and a plane, since 
And 
Based on the perpendicularity of planes, the plane
perpendicular to the plane of the triangle DBC, since it passes through the line AD. We construct the desired angle by dropping the perpendicular from the point WITH to the plane
, let's denote it 
Find the sine of this angle of a right triangle MYSELF. Let us introduce an auxiliary segment BC = a. From a triangle ABC:
From a triangle Navy
(
)
we'll find it.