Theorems establishing a connection between the parallelism of lines and their perpendicularity to the plane. Theorem 2: If two lines are perpendicular to a plane, then they are parallel to each other. Terem 1: If one of two parallel lines is perpendicular to a plane, then the other line is also perpendicular to this plane.
Slide 8 from the presentation "The condition of perpendicularity of a line and a plane". The size of the archive with the presentation is 415 KB.Geometry Grade 10
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Lesson Objectives:
1) to consolidate the questions of theory on the topic "Perpendicularity of a straight line and a plane";
2) to develop skills for solving the main types of problems on the perpendicularity of a straight line and a plane.
During the classes
Report the topic and lesson plan.
II. Actualization of students' knowledge
1) Theoretical survey.
Formulate and prove a theorem on a straight line perpendicular to a plane (prepare at the blackboard for one of the students, then listen to his answer as a whole class).
2) Individual written assignments:
Prove the theorem on the perpendicularity of two parallel lines to a third (1 student);
Prove a theorem establishing a connection between the parallelism of lines and their perpendicularity to the plane (1 student);
Prove a theorem inverse to the theorem establishing a connection between the parallelism of lines and their perpendicularity to the plane (1 student);
Prove the sign of perpendicularity of a line and a plane (1 student).
3) Do-It-Yourself Solution tasks according to ready-made drawings with subsequent verification and discussion, if necessary.
I level: No. 1, 2, 5.
Level II: No. 3, 4, 6.
The point M lies outside the plane ABC.
1. Fig. 1. Prove that the line AC is perpendicular to the plane AMB.
2. Fig. 2. BMDC is a rectangle. Prove: line CD is perpendicular to plane ABC.
3. Fig. 3. ABCD is a rectangle. Prove: AD ⊥ AM.
Solution to problems 1-6.
4. Fig. 4. Prove: BC ⊥ DE.
5. Fig. 5. ABCD - parallelogram. Prove: the line MO is perpendicular to the plane ABC.
6. Fig. 6. ABCD - rhombus. Prove: line BD is perpendicular to plane AMC.
Proof:
AC ⊥ AB (by condition), AC ⊥ AM (by condition),
Proof:
Since BMDC is a rectangle, then ∠MBC = 90°, so
MB ⊥ (ABC) (according to the perpendicularity of the line and the plane).
MB || DC (by the property of the sides of the rectangle). Therefore, DC ⊥ (ABC) (by the theorem on the connection between the parallelism of lines and their perpendicularity to the plane).
Proof:
1) Since ABCD is a rectangle, then ∠ABC = 90°, so BC ⊥ AB, AB ⊂ (ABM)
BC ⊥ (AMV) (according to the perpendicularity of the line and the plane).
2) BC || AD (by the property of the sides of the rectangle). Therefore, AD ⊥ (AMB) (by the theorem on the connection between the parallelism of lines and their perpendicularity to the plane).
3) 
AD ⊥ AM (by definition, a straight line perpendicular to the plane).
No. 4 (Fig. 7)
Proof: Since ΔCMV is isosceles (by condition) and MD is the height, then MD is the median (by the property of the height of an isosceles triangle).
Hence, CD = BD (by definition of the median).
1) Since ΔABC is isosceles (by condition) and AD is the median (by definition), then AD is the height (by the property of the median of an isosceles triangle). Hence, BC ⊥ AD.
2) 
BC ⊥ (AMD) (according to the perpendicularity of the line and the plane).
3) 
BC ⊥ DE (by definition, a straight line perpendicular to the plane).
Proof:
1) AC ∩ BD = O; AO \u003d OS, BO \u003d OD (by the property of the diagonals of a parallelogram).
2) ΔBMD - isosceles (by condition) and MO - median (by definition), therefore, MO - height (by the property of the median of an isosceles triangle).
Therefore, MO ⊥ BD.
3) In ΔAMS: MO ⊥ AC (proved similarly to item 2).
4) 
MO ⊥ (ABC) (according to the perpendicularity of the line and the plane).
No. 6 (Fig. 8)
Proof: AC ⊥ BD and AO = OS, BO = OD (by the property of the diagonals of the rhombus). ΔBMD - isosceles (by condition) and MO - median (by definition), therefore, MO is the height (by the property of the median of an isosceles triangle).
Therefore, MO ⊥ BD.
(on the basis of perpendicularity of a straight line and a plane).
III. Problem solving
The solution is written on the board and in notebooks of problem No. 130 (detailed solution in the textbook), No. 134 (with the help of a teacher), call a strong student to the board.
(Before starting to solve the problem, repeat the concepts: the distance between two points and the distance from a point to a line. Formulate definitions of these concepts.)
Given: ABCD - square; MB - straight 
(Fig. 9).
Find: a) MA, MD, MS; b) ρ (M; AC), ρ (M; BD).
1) AB \u003d BC \u003d CD \u003d AD \u003d n (by the property of the sides of the square).
2) ΔABM and ΔCBM are rectangular, since ∠MBA = ∠MBS = 90°.
According to the Pythagorean theorem: We get,
3) Since BD is the diagonal of a square, then 
4) Since ∠MBA = ∠MBC = 90°, then
MB ⊥ (ABC) (according to the perpendicularity of the line and the plane). Hence, MB ⊥ BD, BD ⊂ (ABC) (by the definition of a line perpendicular to the plane).
5) ΔMBD - rectangular (because MB ⊥ BD, then ∠MBD = 90°). According to the Pythagorean theorem:
6) ρ (M; BD) = MB (by definition of the distance from a point to a line). Hence, ρ (M; BD) = m.
7) AO = OS, BO = OD (according to the property of the diagonals of a square). Because 
then ΔAMC is isosceles (by definition) and MO is the median (by definition), which means MO is the height (according to the property of the median of an isosceles triangle drawn to its base). Therefore, MO ⊥ AS.
