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Ratios, Proportions, and Similarity. Eleanor Roosevelt High School Chin-Sung Lin. ERHS Math Geometry. Ratio and Proportion. Mr. Chin-Sung Lin. ERHS Math Geometry. A ratio is a comparison by division of two quantities that have the same units of measurement

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Ratios proportions and similarity

Ratios, Proportions, and Similarity

Eleanor Roosevelt High School

Chin-Sung Lin


Ratio and proportion

ERHS Math Geometry

Ratio and Proportion

Mr. Chin-Sung Lin


Definition of ratio

ERHS Math Geometry

A ratio is a comparison by division of two quantities that have the same units of measurement

The ratio of two numbers, a and b, where b is not zero, is the number a/b

e.g. AB = 4 cm and CD = 5 cm

AB 4 cm 4

CD 5 cm 5

* A ratio has no units of measurement

Definition of Ratio

or 4 to 5 or 4 : 5

=

=

Mr. Chin-Sung Lin


Definition of ratio1

ERHS Math Geometry

Since a ratio, like a fraction, is a comparison of two numbers by division, a ratio can be simplified by dividing each term of the ratio by a common factor

e.g. AB = 20 cm and CD = 5 cm

AB 20 cm 4

CD 5 cm 1

* A ratio is in simplest form (or lowest terms) when the terms of the ratio have no common factor greater than 1

Definition of Ratio

or 4 to 1 or 4 : 1

=

=

Mr. Chin-Sung Lin


Definition of ratio2

ERHS Math Geometry

A ratio can also be used to express the relationship among three or more numbers

e.g. the measures of the angles of a triangle are 45, 60, and 75, the ratio of these measures can be written as

45 : 60 : 75 or, in lowest terms,

?

Definition of Ratio

Mr. Chin-Sung Lin


Definition of ratio3

ERHS Math Geometry

A ratio can also be used to express the relationship among three or more numbers

e.g. the measures of the angles of a triangle are 45, 60, and 75, the ratio of these measures can be written as

45 : 60 : 75 or, in lowest terms,

3 : 4 : 5

Definition of Ratio

Mr. Chin-Sung Lin


Ratio example

ERHS Math Geometry

If the lengths of the sides of a triangle are in the ratio 3 : 3 : 4, and the perimeter of the triangle is 120 cm, Find the lengths of the sides

Ratio Example

Mr. Chin-Sung Lin


Ratio example1

ERHS Math Geometry

If the lengths of the sides of a triangle are in the ratio 3 : 3 : 4, and the perimeter of the triangle is 120 cm, Find the lengths of the sides

x: the greatest common factor

three sides: 3x, 3x, and 4x

3x + 3x + 4x = 120

10x = 120, x = 12

The measures of the sides: 3(12), 3(12), and 4(12) or 36 cm, 36 cm, and 48 cm

Ratio Example

Mr. Chin-Sung Lin


Definition of rate

ERHS Math Geometry

A rate is a comparison by division of two quantities that have different units of measurement

e.g. A UFO moves 400 m in 2 seconds

the rate of distance per second (speed) is

distance 400 m

time 2 s

* A rate has unit of measurement

Definition of Rate

200 m/s

=

=

Mr. Chin-Sung Lin


Definition of proportion

ERHS Math Geometry

A proportion is a statement that two ratios are equal. It can be read as “a is to b as c is to d”

a c

b d

* When three or more ratios are equal, we can write and extended proportion

a ce g

b df h

Definition of Proportion

=

or a : b = c : d

=

=

=

Mr. Chin-Sung Lin


Definition of extremes and means

ERHS Math Geometry

A proportion a : b = c : d

a : b = c : d

Definition of Extremes and Means

extremes

means

Mr. Chin-Sung Lin


Definition of constant of proportionality

ERHS Math Geometry

When x is proportional to y and y = kx, k is called the constant of proportionality

y

x

Definition of Constant of Proportionality

is the constant of proportionality

k =

Mr. Chin-Sung Lin


Properties of proportions

ERHS Math Geometry

Properties ofProportions

Mr. Chin-Sung Lin


Cross product property

ERHS Math Geometry

In a proportion, the product of the extremes is equal to the product of the means

a c

b d

b ≠ 0, d ≠ 0

* The terms b and c of the proportion are called means, and the terms a and d are the extremes

Cross-Product Property

then a x d = b x c

=

Mr. Chin-Sung Lin


Interchange extremes property

ERHS Math Geometry

In a proportion, the extremes may be interchanged

a c d c

b d b a

b ≠ 0, d ≠ 0, a ≠ 0

Interchange Extremes Property

then

=

=

Mr. Chin-Sung Lin


Interchange means property alternation property

ERHS Math Geometry

Interchange Means Property(Alternation Property)

In a proportion, the means may be interchanged

a c a b

b d c d

b ≠ 0, d ≠ 0, c ≠ 0

then

=

=

Mr. Chin-Sung Lin


Inversion property

ERHS Math Geometry

The proportions are equivalent when inverse both ratios

a c b d

b d a c

b ≠ 0, d ≠ 0, a ≠ 0, c ≠ 0

Inversion Property

then

=

=

Mr. Chin-Sung Lin


Equal factor products property

ERHS Math Geometry

If the products of two pairs of factors are equal, the factors of one pair can be the means and the factors of the other the extremes of a proportion

c a

b d

b ≠ 0, d ≠ 0

a, b are the means, and c, d are the extremes

Equal Factor Products Property

a x b = c x d then

=

Mr. Chin-Sung Lin


Composition property

ERHS Math Geometry

The proportions are equivalent when adding unity to both ratios

a c a + b c + d

b d b d

b ≠ 0, d ≠ 0

Composition Property

then

=

=

Mr. Chin-Sung Lin


Division property

ERHS Math Geometry

The proportions are equivalent when subtracting unity to both ratios

a c a - b c - d

b d b d

b ≠ 0, d ≠ 0

Division Property

then

=

=

Mr. Chin-Sung Lin


Definition of geometric mean mean proportional

ERHS Math Geometry

Definition of Geometric Mean (Mean Proportional)

Suppose a, x, and d are positive real numbers

a x

x d

Then, x is called the geometric mean or mean proportional, between a and d

then x2 = a d or x = √ad

=

Mr. Chin-Sung Lin


Application examples

ERHS Math Geometry

Application Examples

Mr. Chin-Sung Lin


Example ratios and rates

ERHS Math Geometry

Maria has two job opportunities. If she works for a healthcare supplies store, she will be paid $60 daily by working 5 hours a day. If she works for a grocery store, she will be paid $320 weekly by working 8 hours a day and 5 days per week

What is the pay rate for each job?

What is the ratio between the pay rates of healthcare supplies store and grocery store?

Example - Ratios and Rates

Mr. Chin-Sung Lin


Example ratios and rates1

ERHS Math Geometry

Maria has two job opportunities. If she works for a healthcare supplies store, she will be paid $60 daily by working 5 hours a day. If she works for a grocery store, she will be paid $320 weekly by working 8 hours a day and 5 days per week

What is the pay rate for each job?

Healthcare: $12/hr, Grocery: $8/hr

What is the ratio between the pay rates of healthcare supplies store and grocery store?

Healthcare:Grocery = 12:8 = 3:2

Example - Ratios and Rates

Mr. Chin-Sung Lin


Example ratios

ERHS Math Geometry

The perimeter of a rectangle is 48 cm. If the length and width of the rectangle are in the ratio of 2 to 1

What is the length of the rectangle?

Example - Ratios

Mr. Chin-Sung Lin


Example ratios1

ERHS Math Geometry

The perimeter of a rectangle is 48 cm. If the length and width of the rectangle are in the ratio of 2 to 1

What is the length of the rectangle?

Width:x

Length:2x

2 (2x + x) = 48

3x = 24, x = 8

2x = 16

Length is 16 cm

Example - Ratios

Mr. Chin-Sung Lin


Example ratios2

ERHS Math Geometry

The measures of an exterior angle of a triangle and the adjacent interior angle are in the ratio 7 : 3. Find the measure of the exterior angle

Example - Ratios

Mr. Chin-Sung Lin


Example ratios3

ERHS Math Geometry

The measures of an exterior angle of a triangle and the adjacent interior angle are in the ratio 7 : 3. Find the measure of the exterior angle

Exterior angle:7x

Interior angle:3x

7x + 3x = 180

10x = 180, x = 18

7x = 126

Measure of the exterior angle is 126

Example - Ratios

Mr. Chin-Sung Lin


Example proportions

ERHS Math Geometry

Solve the proportions for x

2 x + 2

5 2x - 1

Example - Proportions

=

Mr. Chin-Sung Lin


Example proportions1

ERHS Math Geometry

Solve the proportions for x

2 x + 2

5 2x - 1

2 ( 2x – 1) = 5 (x + 2)

4x – 2 = 5x + 10

-12 = x

x = -12

Example - Proportions

=

Mr. Chin-Sung Lin


Example geometric mean

ERHS Math Geometry

If the geometric mean between x and 4x is 8, solve for x

Example - Geometric Mean

Mr. Chin-Sung Lin


Example geometric mean1

ERHS Math Geometry

If the geometric mean between x and 4x is 8, solve for x

x (4x) = 82

4x2 = 64

x2 = 16

x = 4

Example - Geometric Mean

Mr. Chin-Sung Lin


Example properties of proportions

ERHS Math Geometry

  • 4x2 15y2

  • 3 2

  • 8x2 = 45 y2

  • 4 3

  • 15y2 2x2

  • 5 4x

  • 2x 9y2

If

Example - Properties of Proportions

, which of the following

=

statements are true? Why? (x≠ 0, y ≠ 0)

4x + 9y 5y + 2x

9y 2x

9y 2x

4x - 9y 5y - 2x

=

=

=

=

Mr. Chin-Sung Lin


Example properties of proportions1

ERHS Math Geometry

  • 4x2 15y2

  • 3 2

  • 8x2 = 45 y2

  • 4 3

  • 15y2 2x2

  • 5 4x

  • 2x 9y2

If

Example - Properties of Proportions

, which of the following

=

statements are true? Why? (x≠ 0, y ≠ 0)

4x + 9y 5y + 2x

9y 2x

9y 2x

4x - 9y 5y - 2x

=

=

=

=

Mr. Chin-Sung Lin


Midsegment theorem

ERHS Math Geometry

Midsegment Theorem

Mr. Chin-Sung Lin


Midsegment theorem1

ERHS Math Geometry

A line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of the third side

Given: ∆ABC, D is the midpoint of AC, and

E is the midpoint of BC

Prove: DE || AB, and

DE = ½ AB

Midsegment Theorem

C

E

D

A

B

Mr. Chin-Sung Lin


Midsegment theorem2

ERHS Math Geometry

A line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of the third side

Given: ∆ABC, D is the midpoint of AC, and

E is the midpoint of BC

Prove: DE || AB, and

DE = ½ AB

Midsegment Theorem

C

E

F

D

A

B

Mr. Chin-Sung Lin


Midsegment theorem3

ERHS Math Geometry

A line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of the third side

Given: ∆ABC, D is the midpoint of AC, and

E is the midpoint of BC

Prove: DE || AB, and

DE = ½ AB

Midsegment Theorem

C

E

F

D

A

B

Mr. Chin-Sung Lin


Midsegment theorem4

ERHS Math Geometry

A line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of the third side

Given: ∆ABC, D is the midpoint of AC, and

E is the midpoint of BC

Prove: DE || AB, and

DE = ½ AB

Midsegment Theorem

C

E

F

D

A

B

Mr. Chin-Sung Lin


Midsegment theorem5

ERHS Math Geometry

A line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of the third side

Given: ∆ABC, D is the midpoint of AC, and

E is the midpoint of BC

Prove: DE || AB, and

DE = ½ AB

Midsegment Theorem

C

E

F

D

A

B

Mr. Chin-Sung Lin


Divided proportionally theorem

ERHS Math Geometry

Divided Proportionally Theorem

Mr. Chin-Sung Lin


Definition of divided proportionally

ERHS Math Geometry

Two line segments are divided proportionally when the ratio of the lengths of the parts of one segment is equal to the ratio of the lengths of the parts of the other

e.g., in ∆ABC,

DC/AD = 2/1

EC/BE = 2/1

then,

the points D and E divide AC and BC proportionally

Definition of Divided Proportionally

C

E

D

A

B

Mr. Chin-Sung Lin


Divided proportionally theorem1

ERHS Math Geometry

If two line segments are divided proportionally, then the ratio of the length of a part of one segment to the length of the whole is equal to the ratio of the corresponding lengths of the other segment

Given: In ∆ABC,

AD/DB = AE/EC

Prove: AD/AB = AE/AC

Divided Proportionally Theorem

A

E

D

B

C

Mr. Chin-Sung Lin


Divided proportionally theorem2

A

D

E

B

C

ERHS Math Geometry

StatementsReasons

1. AD/DB = AE/EC1. Given

2. DB/AD = EC/AE 2. Inversion property

3. (DB+AD)/AD = (EC+AE)/AE3. Composition property

4. DB+AD = AB, EC+AE = AC4. Partition postulate

5. AB/AD = AC/AE5. Substitution postulate

6. AD/AB = AE/AC 6. Inversion property

Divided Proportionally Theorem

Mr. Chin-Sung Lin


Converse of divided proportionally theorem

ERHS Math Geometry

If the ratio of the length of a part of one line segment to the length of the whole is equal to the ratio of the corresponding lengths of another line segment, then the two segments are divided proportionally

Given: In ∆ABC,

AD/AB = AE/AC

Prove: AD/DB = AE/EC

Converse of Divided Proportionally Theorem

A

E

D

B

C

Mr. Chin-Sung Lin


Converse of divided proportionally theorem1

A

D

E

B

C

ERHS Math Geometry

StatementsReasons

1. AD/AB = AE/AC 1. Given

2. AB/AD = AC/AE 2. Inversion property

3. (AB-AD)/AD = (AC-AE)/AE3. Division property

4. AB-AD = DB, AC-AE = EC4. Partition postulate

5. DB/AD = EC/AE5. Substitution postulate

6. AD/DB = AE/EC 6. Inversion property

Converse of Divided Proportionally Theorem

Mr. Chin-Sung Lin


Divided proportionally theorem converse of divided proportionally theorem

ERHS Math Geometry

Two line segments are divided proportionally if and only if the ratio of the length of a part of one segment to the length of the whole is equal to the ratio of the corresponding lengths of the other segment

Divided Proportionally Theorem & Converse of Divided Proportionally Theorem

A

E

D

B

C

Mr. Chin-Sung Lin


Application examples1

ERHS Math Geometry

Application Examples

Mr. Chin-Sung Lin


Example divided proportionally

ERHS Math Geometry

In ∆ABC, D is the midpoint of AB and E is the midpoint of AC. BC = 7x+5, DE = 4x-2, BD = 2x+1, AC = 9x+1 Find DE, BC, BD, AB, AC, and AE

Example - Divided Proportionally

A

E

D

B

C

Mr. Chin-Sung Lin


Example divided proportionally1

ERHS Math Geometry

In ∆ABC, D is the midpoint of AB and E is the midpoint of AC. BC = 7x+5, DE = 4x-2, BD = 2x+1, AC = 9x+1 Find DE, BC, BD, AB, AC, and AE

7x + 5 = 2 (4x – 2)

7x + 5 = 8x – 4

x = 9

DE = 4 (9) – 2 = 34

BC = 2 (34) = 68

BD = 2 (9) + 1 = 19

AB = 2 (19) = 38

AC = 9 (9) + 1 = 82

AE = 82/2 = 41

Example - Divided Proportionally

A

E

D

B

C

Mr. Chin-Sung Lin


Example divided proportionally2

ERHS Math Geometry

ABC and DEF are line segments. If AB = 10, AC = 15, DE = 8, and DF = 12,do B and E divide ABC and DEF proportionally?

Example - Divided Proportionally

Mr. Chin-Sung Lin


Example divided proportionally3

ERHS Math Geometry

ABC and DEF are line segments. If AB = 10, AC = 15, DE = 8, and DF = 12,do B and E divide ABC and DEF proportionally?

BC = 15 – 10 = 5

EF = 12 – 8 = 4

AB : BC = 10 : 5 = 2 : 1

DE : EF = 8 : 4 = 2 : 1

AB : BC = DE : EF

So, B and E divide ABC and DEF proportionally

Example - Divided Proportionally

Mr. Chin-Sung Lin


Similar triangles

ERHS Math Geometry

Similar Triangles

Mr. Chin-Sung Lin


Definition of similar figures

ERHS Math Geometry

Two figures that have the same shape but not necessarily the same size are called similar figures

Definition of Similar Figures

Mr. Chin-Sung Lin


Definition of similar polygons

ERHS Math Geometry

  • Two (convex) polygons are similar (~) if their consecutive vertices can be paired so that:

  • Corresponding angles are congruent

  • The lengths of corresponding sides are proportional (have the same ratio, called ratio of similitude)

Definition of Similar Polygons

Mr. Chin-Sung Lin


Definition of similar polygons1

B

A

P

W

X

Q

S

Z

Y

R

D

C

ERHS Math Geometry

  • Both conditions must be true for polygons to be similar:

  • Corresponding angles are congruent

  • The lengths of corresponding sides are proportional

Definition of Similar Polygons

corresponding angles are congruent

6

6

4

60o

60o

6

12

9

corresponding sides are proportional

Mr. Chin-Sung Lin


Definition of similar polygons2

ERHS Math Geometry

If two polygons are similar, then their corresponding angles are congruent and their corresponding sides are in proportion

and

If two polygons have corresponding angles that are congruent and corresponding sides that are in proportion, then the polygons are similar

Definition of Similar Polygons

Mr. Chin-Sung Lin


Definition of similar triangles

ERHS Math Geometry

Triangles are similar if their corresponding angles are equal and their corresponding sides are proportional

(The number represented by the ratio of similitude is called the constant of proportionality)

Definition of Similar Triangles

Mr. Chin-Sung Lin


Example of similar triangles

A

X

10

5

6

3

Z

Y

4

C

B

8

ERHS Math Geometry

A  X, B  Y, C  Z

AB = 6, BC = 8, and CA = 10

XY = 3, YZ = 4 and ZX = 5

Show that ABC~XYZ

Example of Similar Triangles

Mr. Chin-Sung Lin


Example of similar triangles1

A

X

10

5

6

3

Z

Y

4

C

B

8

ERHS Math Geometry

A  X, B  Y, C  Z

AB BC CA 2

XY YZ ZX 1

Therefore ABC~XYZ

Example of Similar Triangles

=

=

=

Mr. Chin-Sung Lin


Example of similar triangles2

?

8

?

4

6

?

ERHS Math Geometry

The sides of a triangle have lengths 4, 6, and 8.

Find the sides of a larger similar triangle if the constant of proportionality is 5/2

Example of Similar Triangles

Mr. Chin-Sung Lin


Example of similar triangles3

8

4

6

ERHS Math Geometry

Assume x, y, and z are the sides of the larger triangle, then

x 5 y 5 z 5

4 2 8 2 6 2

Example of Similar Triangles

=

=

=

y = 20

x = 10

z = 15

Mr. Chin-Sung Lin


Example of similar triangles4

A

X

18

12

9

?

Z

Y

?

B

C

15

ERHS Math Geometry

In ABC, AB = 9, BC = 15, AC = 18.

If ABC~XYZ, and XZ = 12, find XY and YZ

Example of Similar Triangles

Mr. Chin-Sung Lin


Example of similar triangles5

A

X

18

12

9

6

Z

Y

10

B

C

15

ERHS Math Geometry

SinceABC~XYZ, and XZ = 12, then

XY YZ 12

9 15 18

Example of Similar Triangles

=

=

Mr. Chin-Sung Lin


Equivalence relation of similarity

ERHS Math Geometry

Equivalence Relation of Similarity

Mr. Chin-Sung Lin


Reflexive property

ERHS Math Geometry

Any geometric figure is similar to itself

ABC~ABC

Reflexive Property

Mr. Chin-Sung Lin


Symmetric property

ERHS Math Geometry

A similarity between two geometric figures may be expressed in either order

If ABC~DEF, then DEF~ABC

Symmetric Property

Mr. Chin-Sung Lin


Transitive property

ERHS Math Geometry

Two geometric figures similar to the same geometric figure are similar to each other

If ABC~DEF, and DEF~RST, then ABC~RST

Transitive Property

Mr. Chin-Sung Lin


Postulates theorems

ERHS Math Geometry

Postulates & Theorems

Mr. Chin-Sung Lin


Postulate of similarity

ERHS Math Geometry

For any given triangle there exists a similar triangle with any given ratio of similitude

Postulate of Similarity

Mr. Chin-Sung Lin


Angle angle similarity theorem aa

A

X

Z

Y

C

B

ERHS Math Geometry

If two angles of one triangle are congruent to two corresponding angles of another triangle, then the triangles are similar

Given: ABC and XYZ with A X, and C Z

Prove: ABC~XYZ

Angle-Angle Similarity Theorem (AA~)

Mr. Chin-Sung Lin


Angle angle similarity theorem aa1

A

X

R

Z

T

Y

S

C

B

ERHS Math Geometry

StatementsReasons

1. A X, and C Z1. Given

2. Draw RST ~ ABC 2. Postulate of similarity

with RT/AC = XZ/AC

3. R A, TC3. Definition of similar triangles

4. R X, TZ 4. Transitive property

5. AC =AC5. Reflecxive postulate

6. RT = XZ6. Multiplication postulate

7. RST XYZ7. SAS postulate

8. RST ~XYZ8. Congruent ’s are similar ’s

9. ABC ~XYZ9. Transitive property of similarity

Angle-Angle Similarity Theorem (AA~)


Example of aa similarity theorem

E

A

C

45o

45o

D

B

ERHS Math Geometry

Given: mA = 45 and mD = 45

Prove: ABC~DEC

Example of AA Similarity Theorem

Mr. Chin-Sung Lin


Example of aa similarity theorem1

E

A

45o

45o

C

D

B

ERHS Math Geometry

StatementsReasons

1. mA = 45 and mD = 45 1. Given

2. A D 2. Substitution property

3. ACB DCE3. Vertical angles

4. ABC~DEC 4. AA similarity theorem

Example of AA Similarity Theorem

Mr. Chin-Sung Lin


Side side side similarity theorem sss

A

X

Z

Y

C

B

ERHS Math Geometry

Two triangles are similar if the three ratios of corresponding sides are equal

Given: AB/XY = AC/XZ = BC/YZ

Prove: ABC~XYZ

Side-Side-Side Similarity Theorem (SSS~)

Mr. Chin-Sung Lin


Side side side similarity theorem sss1

A

X

Z

Y

C

B

ERHS Math Geometry

Two triangles are similar if the three ratios of corresponding sides are equal

Given: AB/XY = AC/XZ = BC/YZ

Prove: ABC~XYZ

Side-Side-Side Similarity Theorem (SSS~)

E

D

Mr. Chin-Sung Lin


Side side side similarity theorem sss2

A

X

Z

Y

C

B

ERHS Math Geometry

StatementsReasons

1. AB/XY = AC/XZ = BC/YZ1. Given

2. Draw DE, D is on AB, E is on AC2. Postulate of similarity

with AD = XY, DE || BC

3. ADE B, and AED C3. Corresponding angles

4. ADE ~ABC4. AA similaity theorem

5. AB/AD = AC/AE = BC/DE5. Corresponding sides proportional

6. AB/XY = AC/AE = BC/DE6. Substitutin postulate

7. AC/AE = AC/XZ, BC/DE = BC/YZ7. Transitive postulate

8. (AC)(XZ) = (AC)(AE)8. Cross product

(BC)(DE) = (BC)(YZ)

9. AE = XZ, DE = YZ9. Division postulate

10. ADE XYZ10. SSS postulate

11. ADE ~XYZ11. Congruent ’s are similar ’s

12. ABC ~XYZ12. Transitive property of similarity

E

D

Side-Side-Side Similarity Theorem (SSS~)


Side angle side similarity theorem sas

A

X

Z

Y

C

B

ERHS Math Geometry

Two triangles are similar if the ratios of two pairs of corresponding sides are equal and the corresponding angles included between these sides are congruent

Given: A X, AB/XY = AC/XZ

Prove: ABC~XYZ

Side-Angle-Side Similarity Theorem (SAS~)

Mr. Chin-Sung Lin


Side angle side similarity theorem sas1

A

X

Z

Y

C

B

ERHS Math Geometry

Two triangles are similar if the ratios of two pairs of corresponding sides are equal and the corresponding angles included between these sides are congruent

Given: A X, AB/XY = AC/XZ

Prove: ABC~XYZ

Side-Angle-Side Similarity Theorem (SAS~)

E

D

Mr. Chin-Sung Lin


Side angle side similarity theorem sas2

A

X

Z

Y

C

B

ERHS Math Geometry

StatementsReasons

1. A X, AB/XY = AC/XZ1. Given

2. Draw DE, D is on AB, E is on AC2. Postulate of similarity

with AD = XY, DE || BC

3. ADE B, and AED C3. Corresponding angles

4. ADE ~ABC4. AA similaity theorem

5. AB/AD = AC/AE 5. Corresponding sides proportional

6. AB/XY = AC/AE 6. Substitutin postulate

7. AC/AE = AC/XZ7. Transitive postulate

8. (AC)(XZ) = (AC)(AE)8. Cross product

9. AE = XZ9. Division postulate

10. ADE XYZ10. SAS postulate

11. ADE ~XYZ11. Congruent ’s are similar ’s

12. ABC ~XYZ12. Transitive property of similarity

Side-Angle-Side Similarity Theorem (SAS~)

E

D


Example of sas similarity theorem

E

6

?

16

A

C

D

8

10

12

B

ERHS Math Geometry

Prove: ABC~DEC

Calculate: DE

Example of SAS Similarity Theorem

Mr. Chin-Sung Lin


Example of sas similarity theorem1

E

6

5

16

A

C

D

8

10

12

B

ERHS Math Geometry

Prove: ABC~DEC

Calculate: DE

Example of SAS Similarity Theorem

Mr. Chin-Sung Lin


Triangle proportionality theorem side splitter theorem

ERHS Math Geometry

Triangle Proportionality Theorem(Side-Splitter Theorem)

Mr. Chin-Sung Lin


Triangle proportionality theorem

A

D

E

C

B

ERHS Math Geometry

If a line parallel to one side of a triangle intersects the other two sides, then it divides them proportionally

Given: DE || BC

Prove: AD AE

DB EC

Triangle Proportionality Theorem

=

Mr. Chin-Sung Lin


Triangle proportionality theorem1

A

D

E

B

C

ERHS Math Geometry

StatementsReasons

1. BC || DE1. Given

2. A A 2. Reflexive property

3. ABC ADE3. Corresponding angles

4. ABC~ADE 4. AA similarity theorem

5. AB/AD = AC/AE5. Corresp. sides proportional

6. (AB-AD)/AD = (AC-AE)/AE6. Proportional by division

7. AB-AD = DB, AC-AE = EC7. Partition postulate

8. DB/AD = EC/AE8. Substitution postulate

9. AD/DB = AE/EC9. Proportional by Inversion

Triangle Proportionality Theorem

Mr. Chin-Sung Lin


Triangle proportionality theorem2

A

D

E

C

B

ERHS Math Geometry

If a line parallel to one side of a triangle intersects the other two sides, then it divides them proportionally

DE || BC

AD AE

DB EC

AD AE DE

AB AC BC

Triangle Proportionality Theorem

=

=

=

Mr. Chin-Sung Lin


Converse of triangle proportionality theorem

A

D

E

C

B

ERHS Math Geometry

If the points at which a line intersects two sides of a triangle divide those sides proportionally, then the line is parallel to the third side

Given: AD AE

DB EC

Prove: DE || BC

Converse of Triangle Proportionality Theorem

=

Mr. Chin-Sung Lin


Converse of triangle proportionality theorem1

A

D

E

B

C

ERHS Math Geometry

StatementsReasons

1. AD/DB = AE/EC1. Given

2. DB/AD = EC/AE2. Proportional by Inversion

3. (DB+AD)/AD = (EC+AE)/AE3. Proportional by composition

4. DB+AD = AB, EC+AE = AC4. Partition postulate

5. AB/AD = AC/AE5. Substitution postulate

6. A A 6. Reflexive property

7. ABC ~ ADE 7. SAS similarity theorem

8. ABC ADE8. Corresponding angles

9. DE || BC 9. Converse of corresponding angles

Converse of Triangle Proportionality Theorem

Mr. Chin-Sung Lin


Triangle proportionality theorem3

A

D

E

C

B

ERHS Math Geometry

A line is parallel to one side of a triangle and intersects the other two sides if and only if the points of intersection divide the sides proportionally

Triangle Proportionality Theorem

Mr. Chin-Sung Lin


Example of triangle proportionality theorem

A

4

6

8

E

D

?

3

?

C

B

ERHS Math Geometry

Given: DE || BC, AD = 4, BD = 3, AE = 6, DE = 8

Calculate: CE and BC

Example of Triangle Proportionality Theorem

Mr. Chin-Sung Lin


Example of triangle proportionality theorem1

A

4

6

8

E

D

4.5

3

14

C

B

ERHS Math Geometry

Given: DE || BC, AD = 4, BD = 3, AE = 6

Calculate: CE and BC

Example of Triangle Proportionality Theorem

Mr. Chin-Sung Lin


Dilations

ERHS Math Geometry

Dilations

Mr. Chin-Sung Lin


Dilations1

ERHS Math Geometry

A dilation of k is a transformation of the plane such that:

The image of point O, the center of dilation, is O

When k is positive and the image of P is P’, then OP and OP’ are the same ray and OP’ = kOP.

When k is negative and the image of P is P’, then OP and OP’ are opposite rays and P’ = – kOP

Dilations

Mr. Chin-Sung Lin


Dilations2

ERHS Math Geometry

When | k | > 1, the dilation is called an enlargement

When 0 < | k | < 1, the dilation is called a contraction

Under a dilation of k with the center at the origin:

P(x, y) → P’(kx, ky) or Dk(x, y) = (kx, ky)

Dilations

Mr. Chin-Sung Lin


Dilations and similar triangles

A (0, ka)

C (kc, 0)

B (kb, 0)

ERHS Math Geometry

For any dilation, the image of a triangle is a similar triangle

Dilations and Similar Triangles

y

y

X (0, a)

x

x

Z (c, 0)

Y (b,0)

Mr. Chin-Sung Lin


Dilations and similar triangles1

A (0, ka)

C (kc, 0)

B (kb, 0)

ERHS Math Geometry

For any dilation, the image of a triangle is a similar triangle

Dilations and Similar Triangles

y

y

X (0, a)

x

x

Z (c, 0)

Y (b,0)

m XY = – a/b

m XZ = – a/c

m YZ = 0

m AB = – ka/kb = – a/b

m AC = – ka/kc = – a/c

m BC = 0

Mr. Chin-Sung Lin


Dilations and similar triangles2

A (0, ka)

C (kc, 0)

B (kb, 0)

ERHS Math Geometry

For any dilation, the image of a triangle is a similar triangle

Dilations and Similar Triangles

y

y

X (0, a)

x

x

Z (c, 0)

Y (b,0)

XY || AB

XZ || AC

YZ || BC

X A

YB

ZC

ABC ~ ADE

Mr. Chin-Sung Lin


Dilations and angle measures

A (0, ka)

C (kc, 0)

B (kb, 0)

ERHS Math Geometry

Under a dilation, angle measure is preserved

Dilations and Angle Measures

y

y

X (0, a)

x

x

Z (c, 0)

Y (b,0)

Mr. Chin-Sung Lin


Dilations and angle measures1

A (0, ka)

C (kc, 0)

B (kb, 0)

ERHS Math Geometry

Under a dilation, angle measure is preserved

We have proved:

Dilations and Angle Measures

y

y

X (0, a)

x

x

Z (c, 0)

Y (b,0)

X A

YB

ZC

Mr. Chin-Sung Lin


Dilations and midpoint

ERHS Math Geometry

Under a dilation, midpoint is preserved

Dilations and Midpoint

y

y

A (0, ka)

X (0, a)

N

M

x

x

Y (b, 0)

B (kb, 0)

Mr. Chin-Sung Lin


Dilations and midpoint1

ERHS Math Geometry

Under a dilation, midpoint is preserved

Dilations and Midpoint

y

y

A (0, ka)

X (0, a)

N

M

x

x

Y (b, 0)

B (kb, 0)

M(X, Y) = (b/2, a/2)

DK (M) = (kb/2, ka/2)

N(A, B) = (kb/2, ka/2)

N(A, B) = DK (M)

Mr. Chin-Sung Lin


Dilations and collinearity

ERHS Math Geometry

Under a dilation, collinearity is preserved

Dilations and Collinearity

A (ka, kb)

y

Q (ke, kf)

X (a, b)

P (e, f)

B (kc, kd)

Y (c, d)

x

Mr. Chin-Sung Lin


Dilations and collinearity1

ERHS Math Geometry

Under a dilation, collinearity is preserved

Dilations and Collinearity

A (ka, kb)

y

m XP = (b – f) / (a – e)

m PY = (f – d) / (e – c)

XPY are collinear

m XP = m PY

(b – f) / (a – e) = (f – d) / (e – c)

Q (ke, kf)

X (a, b)

P (e, f)

B (kc, kd)

Y (c, d)

x

Mr. Chin-Sung Lin


Dilations and collinearity2

ERHS Math Geometry

Under a dilation, collinearity is preserved

Dilations and Collinearity

A (ka, kb)

y

m XP = (b – f) / (a – e)

m PY = (f – d) / (e – c)

XPY are collinear

m XP = m PY

(b – f) / (a – e) = (f – d) / (e – c)

Q (ke, kf)

X (a, b)

P (e, f)

B (kc, kd)

Y (c, d)

x

m AQ = (kb – kf) / (ka – ke)

= (b – f) / (a – e)

m QB = (kf – kd) / (ke – kc)

= (f – d) / (e – c)

m AQ = m QB

AQB are collinear

Mr. Chin-Sung Lin


Dilations example

ERHS Math Geometry

The coordinates of parallelogram EFGH are E(0, 0), F(3, 0), G(4, 2), and H(1, 2). Under D3, the image of EFGH is E’F’G’H’. Show that E’F’G’H’ is a parallelogram. Is parallelism preserved?

Dilations Example

Mr. Chin-Sung Lin


Dilations example1

ERHS Math Geometry

The coordinates of parallelogram EFGH are E(0, 0), F(3, 0), G(4, 2), and H(1, 2). Under D3, the image of EFGH is E’F’G’H’. Show that E’F’G’H’ is a parallelogram. Is parallelism preserved?

Dilations Example

m E’F’ = 0, m G’H’ = 0

m E’F’ = m G’H’

m F’G’ = 2, m H’E’ = 2

m F’G’ = m H’E’

E’F’G’H’ is a parallelogram

Parallelism preserved

D3 (E) = E’ (0, 0)

D3 (F) = F’ (9, 0)

D3 (G) = G’ (12, 6)

D3 (H) = H’ (3, 6)

Mr. Chin-Sung Lin


Proportional relations among segments related to triangles

ERHS Math Geometry

Proportional Relations among Segments Related to Triangles

Mr. Chin-Sung Lin


Proportional relations of segments

ERHS Math Geometry

  • If two triangles are similar, their corresponding

    • sides

    • altitudes

    • medians, and

    • angle bisectors

  • are proportional

Proportional Relations of Segments

Mr. Chin-Sung Lin


Proportional altitudes

A

C

B

ERHS Math Geometry

If two triangles are similar, the lengths of corresponding altitudes have the same ratio as the lengths of any two corresponding sides

Proportional Altitudes

X

Z

Y

Mr. Chin-Sung Lin


Proportional altitudes1

A

C

B

ERHS Math Geometry

If two triangles are similar, the lengths of corresponding altitudes have the same ratio as the lengths of any two corresponding sides

Proportional Altitudes

X

Z

Y

AA Similarity

Mr. Chin-Sung Lin


Proportional medians

A

C

B

ERHS Math Geometry

If two triangles are similar, the lengths of corresponding medians have the same ratio as the lengths of any two corresponding sides

Proportional Medians

X

Z

Y

Mr. Chin-Sung Lin


Proportional medians1

A

C

B

ERHS Math Geometry

If two triangles are similar, the lengths of corresponding medians have the same ratio as the lengths of any two corresponding sides

Proportional Medians

X

Z

Y

SAS Similarity

Mr. Chin-Sung Lin


Proportional angle bisectors

A

C

B

ERHS Math Geometry

If two triangles are similar, the lengths of corresponding angle bisectors have the same ratio as the lengths of any two corresponding sides

Proportional Angle Bisectors

X

Z

Y

Mr. Chin-Sung Lin


Proportional angle bisectors1

A

C

B

ERHS Math Geometry

If two triangles are similar, the lengths of corresponding angle bisectors have the same ratio as the lengths of any two corresponding sides

Proportional Angle Bisectors

X

Z

Y

AA Similarity

Mr. Chin-Sung Lin


Application problem

ERHS Math Geometry

Two triangles are similar. The sides of the smaller triangle have lengths of 4 m, 6 m, and 8 m. The perimeter of the larger triangle is 63 m. Find the length of the shortest side of the larger triangle

Application Problem

Mr. Chin-Sung Lin


Application problem1

ERHS Math Geometry

  • Two triangles are similar. The sides of the smaller triangle have lengths of 4 m, 6 m, and 8 m. The perimeter of the larger triangle is 63 m. Find the length of the shortest side of the larger triangle

    • 4x + 6x + 8x = 63

    • x = 3.5

    • 4x = 14, 6x = 21, 8x = 28

    • The length of the shortest side is 14 m

Application Problem

Mr. Chin-Sung Lin


Concurrence of the medians of a triangle

ERHS Math Geometry

Concurrence of the Medians of a Triangle

Mr. Chin-Sung Lin


Median of a triangle

A

B

B

A

B

A

C

C

C

ERHS Math Geometry

A segment from a vertex to the midpoint of the opposite side of a triangle

Median of a Triangle

Mr. Chin-Sung Lin


Median of a triangle1

B

A

C

ERHS Math Geometry

If BD is the median of ∆ ABC

then,

AD  CD

Median of a Triangle

D

Mr. Chin-Sung Lin


Definition of centroid

B

Centroid

A

C

ERHS Math Geometry

The three medians meet in the centroid or center of mass (center of gravity)

Definition of Centroid

Mr. Chin-Sung Lin


Theorem about centroid

ERHS Math Geometry

Any two medians of a triangle intersect in a point that divides each median in the ratio 2 : 1

Given: AE and CD are medians of ABC that intersect at P

Prove: AP : EP = CP : DP = 2 : 1

Theorem about Centroid

B

E

D

Centroid

1

1

P

2

2

A

C

Mr. Chin-Sung Lin


Theorem about centroid1

ERHS Math Geometry

Any two medians of a triangle intersect in a point that divides each median in the ratio 2 : 1

Given: AE and CD are medians of ABC that intersect at P

Prove: AP : EP = CP : DP = 2 : 1

Theorem about Centroid

B

E

D

1

Centroid

1

1

P

2

2

2

A

C

Mr. Chin-Sung Lin


Theorem about centroid2

ERHS Math Geometry

StatementsReasons

1. AE and CD are medians1. Given

2. D is the midpoint of AB2. Definition of medians

E is the midpoint of BC

3. AC || DE3. Midsegment theorem

DE = ½ AC

4. AED EAC, CDE DCA4. Alternate interior angles

5. APC ~ EPD 5. AA similarity theorem

6. AC:DE = 2:16. Exchange extremes

7. AP:EP = CP:DP = AC:DE7. Corresp. sides proportional

8. AP:EP = CP:DP = 2:18. Transitive property

Theorem about Centroid

B

E

D

1

1

1

P

2

2

2

A

C

Mr. Chin-Sung Lin


Medians concurrence theorem

ERHS Math Geometry

The medians of a triangle are concurrent

Given: AM, BN, and CL are medians of ABC

Prove: AM, BN, and CL are concurrent

Medians Concurrence Theorem

B

L

M

P

Centroid

A

C

N

Mr. Chin-Sung Lin


Medians concurrence theorem1

ERHS Math Geometry

  • The medians of a triangle are concurrent

  • Given: AM, BN, and CL are medians of ABC

  • Prove: AM, BN, and CL are concurrent

  • Proof:

    • AM and BN intersect at P

    • AP : MP = 2 : 1

    • AM and CL intersect at P’

    • AP’ : MP’ = 2 : 1

    • P and P’ are on AM, and divide that line segment in the ratio 2 : 1

    • Therefore, P = P’ and AM, BN, and CL are concurrent

Medians Concurrence Theorem

B

L

M

P

Centroid

P’

A

C

N

Mr. Chin-Sung Lin


Centroid

ERHS Math Geometry

The centroid divides each median in a ratio of 2:1.

Centroid

B

2

1

1

Centroid

2

2

1

A

C

Mr. Chin-Sung Lin


Ratios proportions and similarity

ERHS Math Geometry

Coordinates of Centroid

The coordinates of the vertices of ΔABC are A(0, 0), B(6, 0), and C(0, 3). (a) Find the coordinates of the centroid P of the triangle. (b) Prove that the centroid divides each median in a ratio of 2:1

Mr. Chin-Sung Lin


Ratios proportions and similarity

ERHS Math Geometry

Coordinates of Centroid

The coordinates of the vertices of ΔABC are A(0, 0), B(6, 0), and C(0, 3). (a) Find the coordinates of the centroid P of the triangle. (b) Prove that the centroid divides each median in a ratio of 2:1

Answer (a): P(2, 1)

Mr. Chin-Sung Lin


Coordinates of centroid

ERHS Math Geometry

If the coordinates of the vertices of a triangle are:

A(x1, y1), B(x2, y2), and C(x3, y3), then

the coordinates of the centroid are:

x1+x2+x3 y1+y2+y3

3 3

Coordinates of Centroid

B

,

P (

)

P

Centroid

A

C

Mr. Chin-Sung Lin


Proportions in a right triangle

ERHS Math Geometry

Proportions in a Right Triangle

Mr. Chin-Sung Lin


Ratios proportions and similarity

ERHS Math Geometry

Projection of a Point on a Line

The projection of a point on a line is the foot of the perpendicular drawn from that point to the line

A

Projection

P

Mr. Chin-Sung Lin


Ratios proportions and similarity

ERHS Math Geometry

Projection of a Segment on a Line

The projection of a segment on a line, when the segment is not perpendicular to the line, is the segment whose endpoints are the projections of the endpoints of the given line segment on the line

A

Projection

B

P

Q

Mr. Chin-Sung Lin


Similar triangles within a right triangle

ERHS Math Geometry

Similar Triangles within a Right Triangle

Mr. Chin-Sung Lin


Identify similar triangles

A

D

C

B

ERHS Math Geometry

The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to each other and to the original triangle

Given: ABC, mB = 90, BD is an altitude

Prove: ABC ~ ADB ~ BDC

Identify Similar Triangles

Mr. Chin-Sung Lin


Prove similar triangles

A

D

C

B

ERHS Math Geometry

BAC  DBC, ACB  BCD, then,

similar triangles: ABC ~ BDC

BAD  CAB, ABD  ACB, then,

similar triangles: ADB ~ ABC

BAD  CBD, ABD  BCD, then,

similar triangles: ADB ~ BDC

So, ABC ~ ADB ~ BDC

Prove Similar Triangles

Mr. Chin-Sung Lin


Identify corresponding sides

A

D

C

B

ERHS Math Geometry

ABC ~ BDCABC ~ ADB

AB AC BC AB AC BC

BD BC DC AD AB BD

BC2 = AC  DCAB2 = AC  AD

ADB ~ BDC

AD AB BD

BD BC CD

BD2 = AD  DC

Identify Corresponding Sides

=

=

=

=

=

=

Mr. Chin-Sung Lin


Right triangle altitude theorem

ERHS Math Geometry

Right Triangle Altitude Theorem

Mr. Chin-Sung Lin


Right triangle altitude theorem part i

ERHS Math Geometry

Right Triangle Altitude Theorem (Part I)

The measure of the altitude drawn from the vertex of the right angle of a right triangle to its hypotenuse is the geometric mean between the measures of the two segments of the hypotenuse.

Given: AD is the altitude

Prove: AD2 = BD  DC

A

C

B

D

Mr. Chin-Sung Lin


Right triangle altitude theorem part ii

ERHS Math Geometry

Right Triangle Altitude Theorem (Part II)

If the altitude is drawn to the hypotenuse of a right triangle, each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to the leg.

Given: AD is the altitude

Prove: AB2 = BC  BD

Prove: AC2 = BC  DC

A

C

B

D

Mr. Chin-Sung Lin


Application examples2

ERHS Math Geometry

Application Examples

Mr. Chin-Sung Lin


Application example 1

A

2

D

y

8

x

C

z

B

ERHS Math Geometry

Given: ABC, mB = 90, BD is an altitude

Solve: x, y, and z

Application Example 1

Mr. Chin-Sung Lin


Application example 11

A

2

D

y

8

x

C

z

B

ERHS Math Geometry

Given: ABC, mB = 90, BD is an altitude

Solve: x, y, and z

x = 4

y = 2√5

z = 4√5

Application Example 1

Mr. Chin-Sung Lin


Application example 2

A

6

D

12

x

y

C

z

B

ERHS Math Geometry

Given: ABC, mB = 90, BD is an altitude

Solve: x, y, and z

Application Example 2

Mr. Chin-Sung Lin


Application example 21

A

6

D

12

x

y

C

z

B

ERHS Math Geometry

Given: ABC, mB = 90, BD is an altitude

Solve: x, y, and z

x = 18

y = 6√3

z = 12√3

Application Example 2

Mr. Chin-Sung Lin


Application example 3

ERHS Math Geometry

Given: ABC, mB = 90, BD is an altitude

Solve: x, y, and z

Application Example 3

A

z

D

6

9

y

C

x

B

Mr. Chin-Sung Lin


Application example 31

ERHS Math Geometry

Given: ABC, mB = 90, BD is an altitude

Solve: x, y, and z

x = 6√3

y = 3√3

z = 3

Application Example 3

A

z

D

6

9

y

C

x

B

Mr. Chin-Sung Lin


Application example 4

ERHS Math Geometry

Given: ABC, mB = 90, BD is an altitude

Solve: w, x, y, and z

Application Example 4

A

y

25

D

w

z

x

C

B

20

Mr. Chin-Sung Lin


Application example 41

ERHS Math Geometry

Given: ABC, mB = 90, BD is an altitude

Solve: w, x, y, and z

w = 15

x = 12

y = 9

z = 16

Application Example 4

A

y

25

D

w

z

x

C

B

20

Mr. Chin-Sung Lin


Application example 5

ERHS Math Geometry

Given: ABC, mB = 90

Prove: AB2 + BC2 = AC2 (Pythagorean Theorem)

(Hint: Apply Triangle Altitude Theorem)

Application Example 5

A

D

C

B

Mr. Chin-Sung Lin


Application example 51

ERHS Math Geometry

Given: ABC, mB = 90

Prove: AB2 + BC2 = AC2 (Pythagorean Theorem)

(Hint: Apply Triangle Altitude Theorem)

AB2 = AC  AD

BC2 = AC  DC

AB2 + BC2

= AC  AD + AC  DC

= AC  (AD + DC)

= AC  AC

= AC2

Application Example 5

A

D

C

B

Mr. Chin-Sung Lin


Pythagorean theorem

ERHS Math Geometry

Pythagorean Theorem

Mr. Chin-Sung Lin


Pythagorean theorem1

ERHS Math Geometry

If a triangle is a right triangle, then the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides (the legs)

Given: ABC, mC = 90

Prove: a2 + b2 = c2

Pythagorean Theorem

c

A

b

a

B

C

Mr. Chin-Sung Lin


Converse of pythagorean theorem

ERHS Math Geometry

If the square of the length of one side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle

Given: a2 + b2 = c2

Prove:ABC is a right triangle

with mC = 90

Converse of Pythagorean Theorem

c

A

b

a

B

C

Mr. Chin-Sung Lin


Converse of pythagorean theorem1

ERHS Math Geometry

c

b

b

Converse of Pythagorean Theorem

StatementsReasons

1. a2 + b2 = c21. Given

2. Draw RST with RT = b, ST = a 2. Create a right triangle

m T = 90

3. ST2 + RT2 = RS23. Pythagorean theorem

4. a2 + b2 = RS2, RS2 = c24. Substitution postulate

5. RS = c5. Root postulate

6. ABC RST 6. SSS postulate

7. C T7. CPCTC

8. mC = 908. Substitution postulate

9. ABC is a right triangle9. Definition of a right triangle

a

a

A

R

B

S

C

T


Pythagorean theorem2

ERHS Math Geometry

A triangle is a right triangle if and only if the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides

ABC, mC = 90

if and only if

a2 + b2 = c2

Pythagorean Theorem

c

A

b

a

B

C

Mr. Chin-Sung Lin


Pythagorean example

ERHS Math Geometry

What is the length of the altitude to the base of an isosceles triangle if the length of the base is 24 centimeters and the length of a leg is 15 centimeters?

Pythagorean Example

A

15

?

B

C

D

24

Mr. Chin-Sung Lin


Pythagorean example1

ERHS Math Geometry

What is the length of the altitude to the base of an isosceles triangle if the length of the base is 24 centimeters and the length of a leg is 15 centimeters?

9 cm

Pythagorean Example

A

15

?

B

C

D

24

Mr. Chin-Sung Lin


Pythagorean example2

ERHS Math Geometry

If a cone has a height of 24 centimeters and the radius of the base is 10 centimeters, what is the slant height of the cone?

Pythagorean Example

A

?

24

C

10

B

Mr. Chin-Sung Lin


Pythagorean example3

ERHS Math Geometry

If a cone has a height of 24 centimeters and the radius of the base is 10 centimeters, what is the slant height of the cone?

26 cm

Pythagorean Example

A

?

24

C

10

B

Mr. Chin-Sung Lin


Pythagorean example4

ERHS Math Geometry

If a right prism has a length of 15 cm, a width of 12 cm and a height of 16 cm, what is the length of AB?

Pythagorean Example

A

?

16

12

B

15

Mr. Chin-Sung Lin


Pythagorean example5

ERHS Math Geometry

If a right prism has a length of 15 cm, a width of 12 cm and a height of 16 cm, what is the length of AB?

25 cm

Pythagorean Example

A

?

16

12

B

15

Mr. Chin-Sung Lin


Pythagorean triples

ERHS Math Geometry

Pythagorean Triples

Mr. Chin-Sung Lin


Pythagorean triples1

ERHS Math Geometry

When three integers can be the lengths of the sides of a right triangle, this set of numbers is called a Pythagorean triple

{3, 4, 5}or{3x, 4x, 5x}

{5, 12, 13} or{5x, 12x, 13x}

{7, 24, 25} or{7x, 24x, 25x}

{8, 15, 17} or{8x, 15x, 17x}

{9, 40, 41} or{9x, 40x, 41x}

Pythagorean Triples

Mr. Chin-Sung Lin


Special right triangles

ERHS Math Geometry

Special Right Triangles

Mr. Chin-Sung Lin


45 45 degree right triangle

ERHS Math Geometry

An isosceles right triangle with 45-45-90-degree angles

45-45-Degree Right Triangle

45o

√ 2

1

A

45o

1

B

C

Mr. Chin-Sung Lin


30 60 degree right triangle

ERHS Math Geometry

A right triangle with 30-60-90-degree angles

30-60-Degree Right Triangle

A

60o

2

1

30o

B

C

√ 3

Mr. Chin-Sung Lin


Special right triangle example

ERHS Math Geometry

An isosceles right triangle with a hypotenuse of 4 cm, what is the length of each leg?

Special Right Triangle Example

4

?

A

B

C

Mr. Chin-Sung Lin


Special right triangle example1

ERHS Math Geometry

An isosceles right triangle with a hypotenuse of 4 cm, what is the length of each leg?

Special Right Triangle Example

4

2√ 2

2√ 2

A

B

C

Mr. Chin-Sung Lin


Special right triangle example2

ERHS Math Geometry

A right triangle with a hypotenuse of 4 cm, and one angle of 30o, what is the length of each leg?

Special Right Triangle Example

A

4

?

30o

B

C

?

Mr. Chin-Sung Lin


Special right triangle example3

ERHS Math Geometry

A right triangle with a hypotenuse of 4 cm, and one angle of 30o, what is the length of each leg?

Special Right Triangle Example

A

4

2

30o

B

C

2√ 3

Mr. Chin-Sung Lin


Triangle angle bisector theorem

ERHS Math Geometry

Triangle Angle Bisector Theorem

Mr. Chin-Sung Lin


Triangle angle bisector theorem1

A

2

1

C

B

D

ERHS Math Geometry

If an angle bisector divides one side of a triangle into two line segments, then these two line segments and the other two sides are proportional

1  2

AB BD

AC DC

Triangle Angle Bisector Theorem

=

Mr. Chin-Sung Lin


Triangle angle bisector theorem2

E

A

2

1

3

C

B

D

ERHS Math Geometry

If an angle bisector divides one side of a triangle into two line segments, then these two line segments and the other two sides are proportional

1  2

AB BD

AC DC

Triangle Angle Bisector Theorem

=

Mr. Chin-Sung Lin


Triangle angle bisector theorem3

E

A

1

2

3

C

B

D

ERHS Math Geometry

StatementsReasons

1. Extend BA and draw EC || AD1. Create similar triangle

2. 1 2 2. Given

3. 2 33. Alternate interior angles

4. 1 E 4. Corresponding angles

5. E 3 5. Transitive property

6. AC = AE 6. Conv. of base angle theorem

7. AB/AE = BD/DC7. Triangle proporatationality

8. AB/AC = BD/DC 8. Substitution property

Triangle Angle Bisector Theorem

Mr. Chin-Sung Lin


Triangle angle bisector theorem example 1

A

2

1

6

4

C

B

D

4

?

ERHS Math Geometry

Triangle Angle Bisector Theorem - Example 1

If 1  2, AB = 6, AC = 4 and BD = 4, DC = ?

Mr. Chin-Sung Lin


Triangle angle bisector theorem example 11

ERHS Math Geometry

Triangle Angle Bisector Theorem - Example 1

If 1  2, AB = 6, AC = 4 and BD = 4, DC = ?

AB BD

AC DC

6 4

4 DC

DC = 8/3

=

A

=

2

1

6

4

C

B

D

4

8/3

Mr. Chin-Sung Lin


Triangle angle bisector theorem example 2

ERHS Math Geometry

Triangle Angle Bisector Theorem - Example 2

If 1  2, AB = 8, DC = 10 and BD = 6, AC = ?

C

10

?

D

6

1

2

A

B

8

Mr. Chin-Sung Lin


Triangle angle bisector theorem example 21

ERHS Math Geometry

Triangle Angle Bisector Theorem - Example 2

If 1  2, AB = 8, DC = 10 and BD = 6, AC = ?

AC DC

AB BD

AC 10

8 6

AC = 40/3

C

=

10

=

40/3

D

6

1

2

A

B

8

Mr. Chin-Sung Lin


Application problems

ERHS Math Geometry

Application Problems

Mr. Chin-Sung Lin


Application problem 1

A

2

1

3x - 2

6

C

B

D

x + 1

3

ERHS Math Geometry

If 1  2, AC = 6, CD = 3, AB = 3x -2, BD = x + 1

Calculate AB

Application Problem 1

Mr. Chin-Sung Lin


Application problem 11

A

2

1

3x - 2

6

C

B

D

x + 1

3

ERHS Math Geometry

If 1  2, AC = 6, CD = 3, AB = 3x -2, BD = x + 1

Calculate AB

x = 4

AB = 10

Application Problem 1

Mr. Chin-Sung Lin


Application problem 2

ERHS Math Geometry

If 1  2, AB = 6, AC = 4, BD = x, DC = x - 1

Calculate BC

Application Problem 2

A

2

1

6

4

C

B

D

x

x - 1

Mr. Chin-Sung Lin


Application problem 21

ERHS Math Geometry

If 1  2, AB = 6, AC = 4, BD = x, DC = x - 1

Calculate BC

x = 3

BC = 5

Application Problem 2

A

2

1

6

4

C

B

D

x

x - 1

Mr. Chin-Sung Lin


Proportions products of line segments

ERHS Math Geometry

Proportions & Products of Line Segments

Mr. Chin-Sung Lin


Key questions

ERHS Math Geometry

  • Proving line segments are in proportion

  • Proving line segments have geometric mean (mean proportional)

  • Proving products of line segments are equal

Key Questions

Mr. Chin-Sung Lin


Steps of proofs

ERHS Math Geometry

  • Forming a proportion from a product

  • Selecting triangles containing the line segments

  • Identifying and proving the similar triangles (Draw lines if necessary)

  • Proving the line segments are proportional

Steps of Proofs

Mr. Chin-Sung Lin


Example in proportions

E

A

C

D

B

ERHS Math Geometry

Given: ABC  DEC

Prove: AC BC

CD CE

Example - In Proportions

=

Mr. Chin-Sung Lin


Example geometric mean2

ERHS Math Geometry

Given: ABC  BDC

Prove: BC is geometric mean between AC and DC

Example - Geometric Mean

A

D

C

B

Mr. Chin-Sung Lin


Example equal product

ERHS Math Geometry

Given: AB || DE

Prove: AD * CE = BE * DC

Example - Equal Product

A

D

C

E

B

Mr. Chin-Sung Lin


Proportionality theorem review

ERHS Math Geometry

Proportionality TheoremReview

Mr. Chin-Sung Lin


Triangle proportionality theorem side splitter theorem1

A

D

E

C

B

ERHS Math Geometry

If a line parallel to one side of a triangle intersects the other two sides, then it divides them proportionally

DE || BC

AD AE

DB EC

AD AE DE

AB AC BC

Triangle Proportionality Theorem(Side-Splitter Theorem)

=

=

=

Mr. Chin-Sung Lin


Triangle angle bisector theorem4

ERHS Math Geometry

If an angle bisector divides one side of a triangle into two line segments, then these two line segments and the other two sides are proportional

1  2

AB BD

AC DC

Triangle Angle Bisector Theorem

A

2

1

=

C

B

D

Mr. Chin-Sung Lin


Similar triangles review

ERHS Math Geometry

Similar Triangles Review

Mr. Chin-Sung Lin


Similar triangles aa

A

X

Z

Y

C

B

ERHS Math Geometry

A  X, B  Y

ABC~XYZ

AB BC CA

XY YZ ZX

Similar Triangles - AA

=

=

Mr. Chin-Sung Lin


Similar triangles sas

A

X

Z

Y

C

B

ERHS Math Geometry

A  X, AB/XY = AC/XZ

ABC~XYZ

AB BC

XY YZ

Similar Triangles - SAS

=

Mr. Chin-Sung Lin


Similar triangles parallel sides shared angle

A

C

B

ERHS Math Geometry

DE || BC

ABC~ADE

A  A, B  ADE, C  AED

AB BC AC

AD DE AE

Similar Triangles - Parallel Sides & Shared Angle

=

=

E

D

Mr. Chin-Sung Lin


Similar triangles parallel sides vertical angles

ERHS Math Geometry

DE || AB

ABC~DEC

A  D, B  E, ACB  DCE

AB BC AC

DE EC DC

Similar Triangles - Parallel Sides & Vertical Angles

E

=

=

A

C

D

B

Mr. Chin-Sung Lin


Similar triangles overlapping triangles

A

D

C

B

ERHS Math Geometry

ABC  BDC

ABC~BDC

ABC  BDC, A  DBC, C  C

AB AC BC

BD BC DC

Similar Triangles - Overlapping Triangles

=

=

Mr. Chin-Sung Lin


Similar triangles angle bisector

ERHS Math Geometry

1  2, (Draw EC || AD, 2  3)

ABD ~ EBC

B  B, 1  E, ADB  ECB

BA BD

AE DC

BA BD

AC DC

Similar Triangles - Angle Bisector

E

=

A

2

1

=

3

B

C

D

Mr. Chin-Sung Lin


Application problems1

ERHS Math Geometry

Application Problems

Mr. Chin-Sung Lin


Application problem 12

ERHS Math Geometry

Given: BC || DE, 1  2

Prove: EC AE

BC AC

Application Problem 1

=

A

E

D

1

2

C

B

Mr. Chin-Sung Lin


Application problem 22

B

F

E

A

C

H

ERHS Math Geometry

Given: AF and BH are angle bisectors, BC = AC

Prove: AH * EF = BF * EH

Application Problem 2

Mr. Chin-Sung Lin


Ratios proportions and similarity

ERHS Math Geometry

Q & A

Mr. Chin-Sung Lin


The end

ERHS Math Geometry

The End

Mr. Chin-Sung Lin


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