- 81 Views
- Uploaded on
- Presentation posted in: General

Ratios, Proportions, and Similarity

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

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

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

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

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

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

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

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

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

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

ERHS Math Geometry

A proportion a : b = c : d

a : b = c : d

Definition of Extremes and Means

extremes

means

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Properties ofProportions

Mr. Chin-Sung Lin

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

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

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

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

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

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

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

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

ERHS Math Geometry

Application Examples

Mr. Chin-Sung Lin

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

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

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

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

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

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

ERHS Math Geometry

Solve the proportions for x

2 x + 2

5 2x - 1

Example - Proportions

=

Mr. Chin-Sung Lin

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

ERHS Math Geometry

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

Example - Geometric Mean

Mr. Chin-Sung Lin

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

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

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

ERHS Math Geometry

Midsegment Theorem

Mr. Chin-Sung Lin

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

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

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

ERHS Math Geometry

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

ERHS Math Geometry

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

ERHS Math Geometry

Divided Proportionally Theorem

Mr. Chin-Sung Lin

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

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

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

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

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

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

ERHS Math Geometry

Application Examples

Mr. Chin-Sung Lin

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

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

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

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

ERHS Math Geometry

Similar Triangles

Mr. Chin-Sung Lin

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

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

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

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

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

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

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

?

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

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

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

A

X

18

12

9

6

Z

Y

10

B

C

15

ERHS Math Geometry

SinceABC~XYZ, and XZ = 12, then

XY YZ 12

9 15 18

Example of Similar Triangles

=

=

Mr. Chin-Sung Lin

ERHS Math Geometry

Equivalence Relation of Similarity

Mr. Chin-Sung Lin

ERHS Math Geometry

Any geometric figure is similar to itself

ABC~ABC

Reflexive Property

Mr. Chin-Sung Lin

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

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

ERHS Math Geometry

Postulates & Theorems

Mr. Chin-Sung Lin

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

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

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, TC3. Definition of similar triangles

4. R X, TZ 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~)

E

A

C

45o

45o

D

B

ERHS Math Geometry

Given: mA = 45 and mD = 45

Prove: ABC~DEC

Example of AA Similarity Theorem

Mr. Chin-Sung Lin

E

A

45o

45o

C

D

B

ERHS Math Geometry

StatementsReasons

1. mA = 45 and mD = 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

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

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

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~)

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

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

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

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

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

ERHS Math Geometry

Triangle Proportionality Theorem(Side-Splitter Theorem)

Mr. Chin-Sung Lin

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

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

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

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

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

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

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

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

ERHS Math Geometry

Dilations

Mr. Chin-Sung Lin

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

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

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

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

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

YB

ZC

ABC ~ ADE

Mr. Chin-Sung Lin

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

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

YB

ZC

Mr. Chin-Sung Lin

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

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

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

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

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

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

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

ERHS Math Geometry

Proportional Relations among Segments Related to Triangles

Mr. Chin-Sung Lin

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

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

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

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

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

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

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

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

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

ERHS Math Geometry

Concurrence of the Medians of a Triangle

Mr. Chin-Sung Lin

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

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

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

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

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

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

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

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

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

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

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

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

ERHS Math Geometry

Proportions in a Right Triangle

Mr. Chin-Sung Lin

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

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

ERHS Math Geometry

Similar Triangles within a Right Triangle

Mr. Chin-Sung Lin

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, mB = 90, BD is an altitude

Prove: ABC ~ ADB ~ BDC

Identify Similar Triangles

Mr. Chin-Sung Lin

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

A

D

C

B

ERHS Math Geometry

ABC ~ BDCABC ~ 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

ERHS Math Geometry

Right Triangle Altitude Theorem

Mr. Chin-Sung Lin

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

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

ERHS Math Geometry

Application Examples

Mr. Chin-Sung Lin

A

2

D

y

8

x

C

z

B

ERHS Math Geometry

Given: ABC, mB = 90, BD is an altitude

Solve: x, y, and z

Application Example 1

Mr. Chin-Sung Lin

A

2

D

y

8

x

C

z

B

ERHS Math Geometry

Given: ABC, mB = 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

A

6

D

12

x

y

C

z

B

ERHS Math Geometry

Given: ABC, mB = 90, BD is an altitude

Solve: x, y, and z

Application Example 2

Mr. Chin-Sung Lin

A

6

D

12

x

y

C

z

B

ERHS Math Geometry

Given: ABC, mB = 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

ERHS Math Geometry

Given: ABC, mB = 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

ERHS Math Geometry

Given: ABC, mB = 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

ERHS Math Geometry

Given: ABC, mB = 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

ERHS Math Geometry

Given: ABC, mB = 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

ERHS Math Geometry

Given: ABC, mB = 90

Prove: AB2 + BC2 = AC2 (Pythagorean Theorem)

(Hint: Apply Triangle Altitude Theorem)

Application Example 5

A

D

C

B

Mr. Chin-Sung Lin

ERHS Math Geometry

Given: ABC, mB = 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

ERHS Math Geometry

Pythagorean Theorem

Mr. Chin-Sung Lin

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, mC = 90

Prove: a2 + b2 = c2

Pythagorean Theorem

c

A

b

a

B

C

Mr. Chin-Sung Lin

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 mC = 90

Converse of Pythagorean Theorem

c

A

b

a

B

C

Mr. Chin-Sung Lin

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. mC = 908. Substitution postulate

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

a

a

A

R

B

S

C

T

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, mC = 90

if and only if

a2 + b2 = c2

Pythagorean Theorem

c

A

b

a

B

C

Mr. Chin-Sung Lin

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

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

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

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

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

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

ERHS Math Geometry

Pythagorean Triples

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Special Right Triangles

Mr. Chin-Sung Lin

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

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

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

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

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

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

ERHS Math Geometry

Triangle Angle Bisector Theorem

Mr. Chin-Sung Lin

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

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

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

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

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

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

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

ERHS Math Geometry

Application Problems

Mr. Chin-Sung Lin

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

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

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

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

ERHS Math Geometry

Proportions & Products of Line Segments

Mr. Chin-Sung Lin

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

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

E

A

C

D

B

ERHS Math Geometry

Given: ABC DEC

Prove: AC BC

CD CE

Example - In Proportions

=

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Given: AB || DE

Prove: AD * CE = BE * DC

Example - Equal Product

A

D

C

E

B

Mr. Chin-Sung Lin

ERHS Math Geometry

Proportionality TheoremReview

Mr. Chin-Sung Lin

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

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

ERHS Math Geometry

Similar Triangles Review

Mr. Chin-Sung Lin

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

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

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

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

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

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

ERHS Math Geometry

Application Problems

Mr. Chin-Sung Lin

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

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

ERHS Math Geometry

Q & A

Mr. Chin-Sung Lin

ERHS Math Geometry

The End

Mr. Chin-Sung Lin