# Geometric mapping of plane, and the motion

The contents of this webpage:

- – the lesson "Geometric mapping of planes" is presented in the example of solving problems 139 - 141;
- – in the worksheets under numbers 142 - 143 of this workbook for mathematics, the motion in geometry is considered;
- – the lesson "Equal rectangles" is explained in the example of the problem 144.

## Geometric mapping of planes

The axial symmetry and the central symmetry are examples of the geometric mapping of the plane.

When a plane is mapped onto itself:

1) every point of the plane is put in correspondence with the unique point of the plane.

2) each point of the plane is placed in correspondence with some point of the plane.

**Definition:**

The motion is the mapping of the plane onto itself, in which the distance between the points is kept.

Examples: the point A lies on the line L,

the point B does not lie on the line L

**a)** draw the point A relative to the line L:

A → A (see Figure a)

Draw the point B relative to line L:

B → B_{1}

1) the segment BB_{1} is perpendicular to the line L

2) the segments OB = OB_{1} are equal (see Figure a)

**b)** draw the points A, B and the segment AB relative to the line L (see Figure b):

A → A_{1}

B → B_{1}

AB → A_{1}B_{1}

**c)** draw the segment AB relative to the line L (see Figure c):

the segments are parallel: AB || A_{1}B_{1}

**d)** draw the segment AB relative to the point O (see Figure d):

A → A_{1} : AO = A_{1}O

B → B_{1} : BO = B_{1}O

***

**Problem 139. **

Given:

L – is the axis of the symmetry,

the straight line *a* is parallel to the line L

*a* || L

*a _{1}* – is the mapping of

*a*

*a → **a _{1}*

Prove:

the straight line *a _{1} * is parallel to the line L

*a _{1}* || L

Proof:

Let us take any two points A and B on the line *a*. We obtain the segment AB *a*, lying on the line *a*. Find the mapping of the segment AB. To get that, we draw perpendiculars to the axis L from the points B and A.

Having measured the distance from the points A or B to the axis L, we postpone the same distance by a compass, i.e.

A → A_{1}

B → B_{1}

We connect the point A_{1} with the point B_{1} and obtain the segment A_{1}B_{1} *a _{1}*, lying on the line

*a*.

_{1}It follows, by the construction:

AO = OA_{1}, BO_{1} = O_{1}B_{1}

Since AB || L , then AO = BO_{1} = O_{1}B_{1 }= OA_{1}

Then OA_{1} = OB_{1}, i.e. the points A_{1} and B_{1} are at an equal distance from the axis L.

Therefore, the line *a _{1}* is parallel to the line L

*a _{1}* || L

***

**Problem 140. **

Given:

*a* is the axis of the symmetry, the line

ABCD is the quadrilateral

Build:

the figure F, to which this quadrilateral is mapped

Construction:

1) draw perpendiculars from points A, B, C, D to the axis *a*.

2) measure the distance with the help of a compass from the point O to the points A and D and postpone them so that the distance between the points A and O, D and O is equal to the distance between the symmetrical distances D_{1}O and A_{1}O.

3) Measure the distance with the help of the compass from the point O_{1} to the points B and C and postpone them so that the distance between the points B and O_{1}, C and O_{1} is equal to the distance between the symmetrical distances B_{1}O_{1} and C_{1}O_{1}.

4) Connect in series the points A_{1}, B_{1}, C_{1}, D_{1}.

By construction, the figure F is the quadrilateral.

***

**Problem 141. **

Given:

L is the axis of the symmetry

the straight line *a* is perpendicular to the line L

*a* L

Prove:

the straight line *a* is mapped onto itself relative to the axis L

*a ** **a*

Proof:

1) any point A *a*, lying on the line *a*, is placed in correspondence with the point A_{1} *a*, also lying on the line *a* (by the definition of the symmetry relative to the line L; *a* L ).

2) any point A_{1} *a*, lying on the line *a*, is placed in correspondence with the point A *a*, lying on the line *a*.

Therefore, the line *a* is mapped onto itself relative to the axis L

*a ** **a*

***

## Motion

**Theorem:**

When moving, the segment is mapped to the segment.

Given:

MN – is the segment

M M_{1}

N N_{1}

Prove:

the segment MN is mapped to the segment M_{1}N_{1}

MN M_{1}N_{1}

Proof:

1) We take a point P MN, lying on the segment MN.

Then MP + PN = MN (*)

By moving the point P is mapped to the point P_{1}

P P_{1}

Since the motion keeps the distance between the points, then

MN = M_{1}N_{1}; MP = M_{1}P_{1}, PN = P_{1}N_{1} (1)

Then it follows from the equalities (*) and (1) that

M_{1}P_{1} + P_{1}N_{1} = M_{1}N_{1} , therefore the point P_{1} lies on the segment M_{1}N_{1}

P_{1} M_{1}N_{1}

Then the segment MN is mapped to the segment M_{1}N_{1}

MN M_{1}N_{1}

2) We verify the second condition.

Any point P_{1} M_{1}N_{1}, lying on the segment M_{1}N_{1}, gives the equality M_{1}P_{1} + P_{1}N_{1} = M_{1}N_{1} (**)

By moving the point P_{1} is translated to the point P

P_{1} P

It follows from (**) and (1) that MP + PN = MN

Therefore, the point P MN lies on the segment MN.

So, by movement, the segment MN is translated into the segment M_{1}N_{1}:

MN M_{1}N_{1}

***

**Consequence 1:**

The triangle is translated to a triangle equal to it.

Given:

the triangle ΔABC

A A_{1}

B B_{1}

C C_{1}

Prove:

the triangle ΔABC is translated into the equal triangle ΔA_{1}B_{1}C_{1}

ΔABC ΔA_{1}B_{1}C_{1}: ΔABC = ΔA_{1}B_{1}C_{1}

Proof:

By the proved theorem, the segment is translated into the equal segment.

Then

1) by the motion, the segment AB is translated into its equal segment A_{1}B_{1} , i.e. AB A_{1}B_{1} , so that AB = A_{1}B_{1}

2) by the motion, the segment AC is translated into its equal segment A_{1}C_{1} , i.e. AC A_{1}C_{1} , so that AC = A_{1}C_{1}

3) by the motion, the segment BC is translated into its equal segment B_{1}C_{1} , i.e. BC B_{1}C_{1} , so that BC = B_{1}C_{1}

Therefore, by the motion, the triangle ΔABC is translated into an equal triangle ΔA_{1}B_{1}C_{1}, i.e.

ΔABC ΔA_{1}B_{1}C_{1}:

ΔABC = ΔA_{1}B_{1}C_{1} (according to three sides)

***

**Consequence 2:**

The straight line is translated into a straight line, the ray is translated into a ray, the angle is translated into an equal angle.

**Problem 142. **

Given:

two straight lines are parallel *a || **b*

*a **a _{1}*

*b **b _{1}*

Prove:

parallelism of the lines, i.e. two straight lines are parallel *a _{1} || *

*b*

_{1}Proof (by contradiction):

Suppose that two lines are not parallel *a _{1} *

*b*, then these lines intersect

_{1}*a*

_{1}∩*b*

_{1}Each point of the lines *a* and *b* is given a unique point of the lines *a _{1}* and

*b*.

_{1}Each point of the lines *a _{1}* and

*b*is placed at some point of the straight lines

_{1}*a*and

*b*.

Then the point of the intersection of the lines *a _{1}* and

*b*is set at the intersection point of the lines

_{1}*a*and

*b*, which contradicts the condition of the problem.

Therefore, the two lines are parallel *a _{1} || *

*b*

_{1}***

**Problem 143. **

Given:

O is the point of the central symmetry

*b* is the straight line

Build:

the straight line *b _{1}*, to which the line

*b*is mapped relative to the point O

*b **b _{1}*

Construction:

1) Take any two points, for example, A and B, on the line *b*.

2) Draw the ray BO.

3) Draw the ray AO.

4) Let us measure the segment AO with the help of the compass and on the continuation of the ray AO let us postpone from the point O the distance equal to the distance between the points A and O.

5) Let us measure the segment BO with the help of the compass and, on the continuation of the ray BO, postpone the distance equal to the distance between the points B and O from the point O.

6) We obtain two equal segments AB = A_{1}B_{1}

7) We draw the line *b _{1}* through A

_{1}B

_{1}, then we obtain a straight line

*b*, to which the line

_{1}*b*is mapped relative to the point O

*b **b _{1}*

***

**Problem 144. **

Given:

ABCD is the rectangle

A_{1}B_{1}C_{1}D_{1} is the rectangle

a) AB = A_{1}B_{1}, BC = B_{1}C_{1}

b) A_{1}B_{1} = AB

AC = A_{1}C_{1} are the diagonals

Prove: ABCD = A_{1}B_{1}C_{1}D_{1}

Proof:

a) Let us prove it by contradiction.

Suppose that ABCD ≠ A_{1}B_{1}C_{1}D_{1}

Then AB ≠ A_{1}B_{1} and BC ≠ B_{1}C_{1}, which contradicts this condition.

Therefore, the rectangles are equal ABCD = A_{1}B_{1}C_{1}D_{1}

***

b) We prove by contradiction.

Suppose that ABCD ≠ A_{1}B_{1}C_{1}D_{1}

Then AB ≠ A_{1}B_{1} and BC ≠ B_{1}C_{1}

Since the rectangles are not equal, then the diagonals will not be equal, which contradicts the condition of the problem.

Therefore, the rectangles are equal ABCD = A_{1}B_{1}C_{1}D_{1}

***