# Parallel transfer, rotation of the plane, and similar triangles

The contents of this webpage:

- – the lesson "Parallel transfer" is presented in the example of solving problems 145 - 148;
- – in the worksheets under numbers 149 - 154 of this workbook for mathematics, the rotation of the plane around the point by an angle is considered;
- – revising of the geometry of grade 9 in solutions is given in the examples 155 - 173:

- the angles of the triangle,

- the area of the triangle through the legs and the hypotenuse,

- the calculation of the radius of the circumscribed circle,

- the side of the rhombus,

- the similar triangles.

## Parallel transfer

**Definition:**

**The parallel transfer** onto a vector is a mapping of the plane onto itself, under which each point M is mapped to the point M_{1}, such that two vectors are equal.

=

***

**Problem 145. **

Given:

the vector

the segment AB

Build:

AB → A_{1}B_{1}

A → A_{1} : =

B → B_{1} : =

***

**Theorem****:**

With the parallel transfer to the vector the distance between the points is kept, i.e. **the parallel transfer is the motion**.

Given:

f is the parallel transfer to the vector

M M_{1}

NN_{1}

Prove:

f is the motion (MN = M_{1}N_{1})

Proof:

The point M is translated by the motion to the point M_{1} with the condition that the two vectors are equal: M M_{1}: = MM_{1}

The point N is translated by the motion to the point N_{1} with the condition that the two vectors are equal: NN_{1}: = NN_{1}

Therefore, the obtained segments are parallel MM_{1} || NN_{1} and the constructed segments are equal MM_{1} = NN_{1}

Therefore, the quadrilateral MM_{1}N_{1}N is the parallelogram.

Therefore MN = M_{1}N_{1}, then f is the motion.

***

**Problem 146. **

a)

the triangle ΔABC

A A_{1}:

=

B B_{1}:

=

C C_{1}:

=

b)

the triangle ΔABC

A A_{1}: =

B B_{1}:

=

C C_{1}:

=

***

**Problem 147. **

Given:

the triangle ΔABC

AB = BC

the point D lies on AC: DAC

the point C lies on AD: CAD

BC B_{1}D

a) Build: B_{1}D

b) Prove: ABB_{1}D is the isosceles trapezoid

a)

Construction:

1) draw the line *a* from the point B parallel to the vector : *a* ||

2) the point B is translated by the motion to the point B_{1}

=

3) draw the line B_{1}D parallel to the segment BC:

B_{1}D || BC

***

b)

Proof:

Consider the quadrilateral BB_{1}DC.

Since the bases of BB_{1} || CD and the lateral sides BC || BD are parallel, then BB_{1}DC is the parallelogram (by the definition)

By the property of the parallelogram:

the bases BB_{1} = CD and the lateral sides BC = BD are equal, but AB = BC, then AB = B_{1}D

Since BB_{1} || AD are parallel and AB B_{1}D are not parallel, therefore ABB_{1}D is the trapezoid (by the definition).

Since AB = B_{1}D, then ABB_{1}D is the trapezoid (by the definition).

***

**Problem 148. **

Given:

the triangle ΔABC

EFPQ is the trapezoid

the circle (O; R)

the vector

Build:

circle (O;R) circle (O_{1};R_{1})

ΔABC ΔA_{1}B_{1}C_{1}

EFPQ E_{1}F_{1}P_{1}Q_{1}

Construction:

as it is shown on the picture.

***

## Rotation of the plane around the point by the angle

**Definition:**

**Rotating the plane around the point O by an angle α** is a mapping of the plane onto itself such that each point M is mapped onto the point M

_{1}, such that the rotation angle

MOM_{1} = α and OM_{1} = OM.

O is the center of the rotation

*α* is the angle of the rotation

***

**Problem 149.**

Given:

α = 75° (counter-clockwise)

O is the center of the rotation

AB is the segment

Build:

A_{1}B_{1}

Construction:

1) AA_{1};

AOA_{1} = 75°

OA = OA_{1}

2) BB_{1};

BOB_{1} = 75°

OB = OB_{1}

***

**Theorem:**

The rotation is the motion.

Given:

f is the rotation

α is the angle of the rotation (counter-clockwise)

the point O is the center of the rotation

MN is the segment

Prove:

f is the motion, MN = M_{1}N_{1}

Proof:

Let M→M_{1}; N→N_{1}

Then the triangles are equal ΔOMN = ΔOM_{1}N_{1} by two sides and the angle between them:

OM = OM_{1}

ON = ON_{1}

MON = M_{1}ON_{1}

Then MN = M_{1}N_{1}, therefore, f is the motion..

***

**Problem 150.**

Given:

the point O is the center of the rotation

α = 180°

AB is the segment

AB→A_{1}B_{1}

Build:

A_{1}B_{1}

Construction:

1) AA_{1};

AOA_{1} = 180°

OA = OA_{1}

2) BB_{1};

BOB_{1} = 180°

OB = OB_{1}

***

**Problem 151.**

Given:

the triangle ΔABC

the point A is the center of the rotation

α = 160° (counter-clockwise)

ΔABC→ΔAB_{1}C_{1}

Build: ΔAB_{1}C_{1}

Construction:

1) BB_{1};

BAB_{1} = 160°

BA = B_{1}A

2) CC_{1};

CAC_{1} = 160°

CA = AC_{1}

***

**Problem 152.**

Given:

the point O is the center of the rotation

α = 120°

AB is the segment

AB→A_{1}B_{1}

Build:

A_{1}B_{1}

Construction:

1) AA_{1};

AOA_{1} = 120°

OA = OA_{1}

2) BB_{1};

BOB_{1} = 120°

OB = OB_{1}

***

**Problem 153.**

Given:

the point C is the center of the circle (C; R)

the point O is the center of the rotation

the rotation angle α = 60° (counter-clockwise)

a) the point C and the point O do not coincide

b) the point C and the point O coincide

Build:

circle (C_{1}; R)

Construction:

a)

1) we draw the ray CO

2) CC_{1};

COC_{1} = 60°

CO = C_{1}O

b)

Since the point O as the center of rotation and the point C as the center of the circle coincide, then the circles (C;R) and (C_{1};R) will also coincide.

***

**Problem 154.**

Given:

Δ ABC is the isosceles, equilateral triangle

D is the point of intersection of the bisectors

D is the center of the rotation

the rotation angle α = 120°

Prove:

ΔABCΔABC

Proof:

Since Δ ABC is the regular triangle, then all angles of it are equal to 60°.

Since the point D is the center of the circumscribed and inscribed circle, then

AD = BD = DC = R.

Δ ABD = Δ BDC = Δ DAC (by three sides).

Therefore, that ADB =BDC =CDA

Therefore

AB

BC

CA

I.e. ΔABC → ΔBCA.

So, Δ ABC is mapped to itself.

***

**Revising.**

**Problem 155.**

Given:

the triangle ΔABC

the ratio of the angles

ABC :BCA :CAB = 3 : 7 : 8

Find: the largest angle of the triangle

Solution:

Let *x* be the coefficient of the proportionality. Knowing that the sum of the angles in the triangle is 180°, we compose and solve the equation:

3x + 7x + 8x = 180

18x = 180

x = 10

The largest angle is CAB = 8 • 10 = 80°

Answer: 80°.

***

**Problem 156.**

Given:

the triangle ΔABC is the isosceles triangle,

one angle is greater than the other:

ABC >BAC by 60°

Find: the angle at the base of the triangle

Solution:

Let *x*° be the angle at the base of the triangle. Knowing that the sum of the angles in the triangle is 180°, we will compose and solve the equation:

(x + 60°) + x + x = 180°

3x = 180 – 60

3x = 120

x = 40

Therefore, BAC = 40°.

Answer: 40°.

***

**Problem 157.**

Given:

the triangle ΔABC is right

*c* = 26 cm is the hypotenuse

the ratio of the legs:

*a : **b* = 5 : 12

Find: the greater leg *b*

Solution:

Let *x* be the coefficient of the proportionality. By the theorem of Pythagoras we compose and solve the equation:

(5x)^{2} + (12x)^{2} = 26^{2}

25x^{2} + 144x^{2} = 676

169x^{2} = 676

x^{2} = 4

x = 2

b = 12 • 2 = 24 (cm)

Answer: 24 cm.

***

**Problem 158.**

Given:

the triangle ΔABC,

C = 90°

*b *= 5 is the leg

*c* = 13 is the hypotenuse

Find: the area of the triangle S_{Δ}_{ABC} = ?

Solution:

By the Pythagorean theorem we obtain:

*a* ==== 12

Then the area of the triangle

S_{Δ}_{ABC} = • *ab* = =

= 30 (square units)

Answer: 30 square units.

***

**Problem 159.**

Given:

the triangle ΔABC is isosceles,

C = 90°

*c* = 4 is the hypotenuse

Find: the area of the triangle S_{Δ}_{ABC} = ?

Solution:

S_{Δ}_{ABC} = • *ab*

Since Δ ABC is isosceles, then the angles at the base of 45° and the legs are *a = **b*.

By the Pythagorean theorem we obtain:

*c ^{2}* =

*a*

^{2}+*b*=

^{2}*a*

^{2}+*a*= 2

^{2}*a*

^{2}Then (4)^{2} = 2*a ^{2}*

* a ^{2}* = 16

*a *= 4 (unit)

Then the area of the triangle

S_{Δ}_{ABC} = • *ab* = =

= 8 (square units)

Answer: 8 square units.

***

**Problem 160.**

Given:

the triangle ΔABC,

A = 90°

AH is the median

*a *= 6

*b* = 8

Find: the radius of the circumscribed circle R = ?

Solution:

Since AH is the median, then CH = *c*

By the Pythagorean theorem we obtain:

*c ^{2}* =

*a*

^{2}+*b*

^{2}*c ^{2}* = 36 + 64

*c* = 10 (units)

Then CH = *c* = = 5 (units)

The point H is the center of the circumscribed circle

AH = BH = CH = R

Since R = AH, then R = AH = CH = 5 units.

Answer: 5 units.

***

**Problem 161.**

Given:

the triangle ΔABC,

C = 90°

the ratio of the acute angles

ABC : CAB = 1 : 2

AC = 4

Find: the radius of the circumscribed circle R = ?

Solution:

Let *x* be the coefficient of the proportionality. Knowing that the sum of the angles in the triangle is 180°, we will compose and solve the equation:

x + 2x + 90 = 180

3x = 90

x = 30

Then CAB = 30°,

ABC = 2 • 30° = 60°

Therefore, BC = AB

By the Pythagorean theorem we obtain:

*AC ^{2} + *

*BC*

^{2}=*AB*

^{2}*AC ^{2} + *

*= AB*

^{2}*AC ^{2} = *

*AB*

^{2}*AB ^{2} = =* 64

AB = 8 (units)

R = AD = BD = 8 : 2 = 4 (units)

Answer: 4 units.

***

**Problem 162.**

Given:

the triangle ΔABC,

C = 90°

BC = 3

the radius of the circumscribed circle

R = 2,5

Find: AC = ?

Solution:

R = AH = BH = 2,5

Then AB = 2,5 • 2 = 5

By the Pythagorean theorem we obtain:

AC = = = = 4 (units)

Answer: 4.

***

**Problem 163.**

Given:

the triangle ΔABC,

C = 90°

tg A = 0,6

BC = 3

Find: AC = ?

Solution:

tgA =

0,6 = ; AC = 3 • = 5 (units)

Answer: 5.

***

**Problem 164.**

Given:

the triangle ΔABC,

A = 90°

AH = AC

Find: ABC = ?

Solution:

Since AH = AC, then Δ AHC is isosceles.

The point H is the radius of the inscribed circle, so AH = CH, but AH = AC, therefore, AH = CH = AC.

Then Δ AHC is equilateral.

Therefore, HAC = AHC = HCA = 60°.

ABC = 180° – (90° + 60°) = 30°.

Answer: 30°.

***

**Problem 165.**

Given:

the triangle Δ ABC is regular, equilateral,

the area of the triangle

S_{Δ}_{ABC} = square units

Find: the length of the bisector BH = ?

Solution:

Since Δ ABC – is regular, then all angles are equal to 60°.

Consider Δ ABC is isosceles, where

BAC = BCA = 60°.

Then BH is the median, the height.

Therefore, the segments are perpendicular: BH AC.

Consider the triangles Δ ABH and Δ BHC.

AB = BC, by the given hypothesis.

AH = CH, BH is the median.

BH is the common side.

Therefore, the triangles are equal Δ ABH = Δ BHC.

The area of the triangle is S_{Δ}_{ABC }= 2S_{Δ}_{ABH}

I.e. S_{Δ}_{ABH} = S_{Δ}_{ABC} = • = (square units)

S_{Δ}_{ABH } = AH • BH

Consider the triangle Δ ABH.

Since BH is the bisector, then the angle ABH = 30°, therefore

AH = AB

S_{ΔABH } = AB • BH =

AB • BH = (*)

By the Pythagorean theorem we obtain:

AB^{2} = AH^{2} + BH^{2}

AB^{2} = AB^{2} + BH^{2}

BH^{2} = AB^{2}

BH = AB (**)

Using the result (**) in equation (*), we obtain

AB • AB =

AB^{2} =

AB =

Then AB • BH = • BH =

BH = 1 (unit)

Answer: BH = 1 unit

***

**Problem 166.**

Given:

the triangle Δ ABC is regular, equilateral,

the radius of the circumscribed circle

R =

Find: the area of the triangle

S_{Δ}_{ABC} = ?

Solution:

Consider Δ ABO (AO = BO = R) Δ ABO is the isosceles triangle.

We draw the height OH from the vertex O to AB.

Consider Δ AOH, where AHO = 90°.

Since HAO = 30°, then OH = AO OH = R

OH = • =

By the Pythagorean theorem we obtain:

OH^{2} + AH^{2} = OA^{2}

OH^{2} + AH^{2} =R^{2}

+ AH^{2} = ()^{2} + AH^{2} =

=

AH^{2} = – = AH = =

Then the area of the triangle

S_{Δ}_{AOH} = AH • OH = ••= =

Therefore, S_{Δ}_{ABO} = 2 • S_{Δ}_{AOH} = 2 • = (square units)

Then the area of the triangle

S_{Δ}_{ABC} = 3 • S_{Δ}_{ABO} = 3 • = = 2= 2,25 (square units)

Answer: 2,25 square units

***

**Problem 167.**

Given:

the rhombus area is S_{ABCD} = 384

the ratio of the diagonals of the rhombus:

AC : BD = 3 : 4

Find: the rhombus side AB = ?

Solution:

the area of the rhombus

S_{ABCD} = AC • BD

Let *x* be the coefficient of the proportionality. Then

S_{ABCD} = 3x • 4x

384 = 6x^{2}

x^{2} = 64

x = 8

Therefore, the diagonal BD = 4x = 4 • 8 = 32

AC = 3x = 3 • 8 = 24

AC = 2AO

BD = 2BO

Therefore, half the diagonal AO =AC = • 24 = 12

BO =BD = • 32 = 16

By the Pythagorean theorem we obtain:

AO^{2} + BO^{2} = AB^{2}

The side of the rhombus is AB = == 20

Answer: 20.

***

**Problem 168.**

Given:

the triangle Δ ABD is isosceles,

the base AD = 16

the side AB = 10

Find: the area of the triangle

S_{Δ}_{ABD} = ?

Solution:

the area of the triangle

S_{Δ}_{ABD} = AD • BH

We draw the height BH to the base AD.

By the property of an isosceles triangle:

BH is the median, the bisector, the height.

Since BH is the median, then AH = DH = 16 : 2 = 8 (unit)

Consider the triangle Δ ABH, where the angle AHB = 90°.

By the Pythagorean theorem we obtain:

AB^{2} = AH^{2} + BH^{2}

BH = = == 6 (unit)

Then the area of the triangle

S_{Δ}_{ABD} = AD • BH = •16 • 6 = 48 (square units)

Answer: the area of the triangle is S_{Δ}_{ABD} = 48 square units

***

**Problem 169.**

Given:

the triangle Δ ABC is isosceles,

the height BH = 15

the base AC is greater than the height BH by 15: AC > BH by 15

Find: the base AC = ?

Solution:

Since the triangle Δ ABC is isosceles, then BH is the height, the median, the bisector.

Therefore AH = CH.

Then AC = AH + CH = AH + AH = 2 AH

Consider Δ ABH is the right triangle.

Let AC = (*x*) unit AH = () unit

Then AB = (*x* – 15) unit (by the given condition).

By the Pythagorean theorem, we solve the equation:

(x – 15)^{2} = ()^{2} + 15^{2 }x^{2} – 30x + 225 = + 225

4 (x^{2} – 30x) = x^{2}

4x^{2} – 120x = x^{2}

3x^{2} – 120x = 0 | : x

3x = 120

x = 40

So, 40 units – is the length of the base.

Answer: AC = 40 units

***

## Similar triangles

**Problem 170. **

Given:

the triangle Δ ABC, two angles

A = 54°

B = 18°

CH is the bisector of the angle C

Prove: the similarity of triangles

Δ BHC Δ ABC

Proof:

C = 180° – (A +B)

C = 180° – (54° + 18°) = 108°

Since CH is the bisector of the angle C, then

the angles are equal to

BCH = HCA = 108° : 2 = 54°

Consider Δ BHC

HBC = B = 18°

BCH = A = 54°

Then CHB = C = 108°

Therefore the triangles are similar Δ BHC Δ ABC.

***

**Problem 171. **

Given:

ABCD is the trapezoid,

the top base BC = 4 cm

the bottom base AD = 10 cm

the diagonal BD = 8 cm

Find:

the part of the diagonal BO = ?

the ratio of the perimeters of the triangles

= ?

Solution:

The angles are equal CBO = ODA as the crosswise angles at the parallel lines BC and AD and the secant BD.

The angles are equal BCO = OAD as the crosswise angles at the parallel lines BC and AD and the secant AC.

Then the triangles are similar Δ BCO Δ AOD.

= = = =

= . Then 4AO = 10BO BO = AO

= = 0,4 = k

Let BO = x, AO = 8 – x. Then 10x = 4 • (8 – x)

10x = 32 – 4x

14x = 32

x = 2 (cm)

Therefore, BO = 2cm.

= k = 0,4

Answer: BO = 2cm, = 0,4.

***

**Problem 172. **

Given:

the similar triangles

ΔABC ΔA_{1}B_{1}C_{1 },

AB = 12 dm,

BC = 16 dm,

AC = 20 dm,

the perimeter of the triangle:

P (ΔA_{1}B_{1}C_{1}) = 60 dm

Find:

the sides of the triangle ΔA_{1}B_{1}C_{1}

A_{1}B_{1}, A_{1}C_{1}, B_{1}C_{1}= ?

Solution:

the perimeter of the triangle

P (ΔABC) = 12 +16 + 20 = 48 (dm)

Since the triangles are similar, then

==

=== k (*)

Then the ratio of the perimeters of the triangles

= k (**)

From the equalities (*) and (**) it follows

=

=

B_{1}C_{1} = = 20 (dm)

Then =

=

A_{1}B_{1} = = 15 (dm)

A_{1}C_{1} = P(ΔA_{1}B_{1}C_{1}) – B_{1}C_{1} – A_{1}B_{1} = 60 – 20 – 15 = 25 (dm).

Answer: A_{1}C_{1} = 25 dm, A_{1}B_{1} = 15 dm, B_{1}C_{1} = 20 dm.

***

**Problem 173. **

Given:

ABCD is the trapezoid,

the sides of the trapezoid intersect at the point M:

AB ∩ CD = M,

BC = 5 cm,

AD = 8 cm,

AB = 3,9 cm,

CD = 3,6 cm

Find:

MB, MC = ?

Solution:

Consider the triangles ΔAMD and ΔBMC:

BAD =MBC, as the corresponding angles at the parallel lines BC and AD and the secant AB.

MCB =MDA, as the corresponding angles at the parallel lines BC and AD and the secant CD.

Then, according to the first theorem of the similarity of the triangles:

the triangles are similar Δ AMD Δ BMC.

Therefore,

= =

=,

but AM = AB + BM = 3,9 + BM

Then

8 • BM = 5 (3,9 + BM)

8BM – 5BM = 19,5

3BM = 19,5

BM = 6,5 (cm)

=,

but MD = CD + MC = 3,6 + MC

8 • MC = 5 (3,6 + MC)

3MC = 18

MC = 6 (cm)

Answer: 6,5 cm, 6 cm.

***