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Ratios, Proportions, and Similarity PowerPoint Presentation

Ratios, Proportions, and Similarity

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Presentation Transcript

### Definition of Ratio

### Definition of Ratio

### Definition of Ratio

### Definition of Ratio

### Ratio Example

### Ratio Example

### Definition of Rate

### Definition of Proportion

### Definition of Extremes and Means

### Definition of Constant of Proportionality

### Cross-Product Property

### Interchange Extremes Property

### Interchange Means Property(Alternation Property)

### Inversion Property

### Equal Factor Products Property

### Composition Property

### Division Property

### Definition of Geometric Mean (Mean Proportional)

### Example - Ratios and Rates

### Example - Ratios and Rates

### Example - Ratios

### Example - Ratios

### Example - Ratios

### Example - Ratios

### Example - Proportions

### Example - Proportions

### Example - Geometric Mean

### Example - Geometric Mean

### Example - Properties of Proportions

### Example - Properties of Proportions

### Midsegment Theorem

### Midsegment Theorem

### Midsegment Theorem

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

### Midsegment Theorem

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

### Midsegment Theorem

### Definition of Divided Proportionally

### Divided Proportionally Theorem

### Divided Proportionally Theorem

### Converse of Divided Proportionally Theorem

### Converse of Divided Proportionally Theorem

### Divided Proportionally Theorem & Converse of Divided Proportionally Theorem

### Example - Divided Proportionally

### Example - Divided Proportionally

### Example - Divided Proportionally

### Example - Divided Proportionally

### Definition of Similar Figures

### Definition of Similar Polygons

### Definition of Similar Polygons

### Definition of Similar Polygons

### Definition of Similar Triangles

### Example of Similar Triangles

### Example of Similar Triangles

### Example of Similar Triangles

### Example of Similar Triangles

### Example of Similar Triangles

### Example of Similar Triangles

### Reflexive Property

### Symmetric Property

### Transitive Property

### Postulate of Similarity

### Angle-Angle Similarity Theorem (AA~)

### Angle-Angle Similarity Theorem (AA~)

### Example of AA Similarity Theorem

### Example of AA Similarity Theorem

### Side-Side-Side Similarity Theorem (SSS~)

### Side-Side-Side Similarity Theorem (SSS~)

### Side-Side-Side Similarity Theorem (SSS~)

### Side-Angle-Side Similarity Theorem (SAS~)

### Side-Angle-Side Similarity Theorem (SAS~)

### Side-Angle-Side Similarity Theorem (SAS~)

### Example of SAS Similarity Theorem

### Example of SAS Similarity Theorem

### Triangle Proportionality Theorem

### Triangle Proportionality Theorem

### Triangle Proportionality Theorem

### Converse of Triangle Proportionality Theorem

### Converse of Triangle Proportionality Theorem

### Triangle Proportionality Theorem

### Example of Triangle Proportionality Theorem

### Example of Triangle Proportionality Theorem

### Dilations

### Dilations

### Dilations and Similar Triangles

### Dilations and Similar Triangles

### Dilations and Similar Triangles

### Dilations and Angle Measures

### Dilations and Angle Measures

### Dilations and Midpoint

### Dilations and Midpoint

### Dilations and Collinearity

### Dilations and Collinearity

### Dilations and Collinearity

### Dilations Example

### Dilations Example

### Proportional Relations of Segments

### Proportional Altitudes

### Proportional Altitudes

### Proportional Medians

### Proportional Medians

### Proportional Angle Bisectors

### Proportional Angle Bisectors

### Application Problem

### Application Problem

### Median of a Triangle

### Median of a Triangle

### Definition of Centroid

### Theorem about Centroid

### Theorem about Centroid

### Theorem about Centroid

### Medians Concurrence Theorem

### Medians Concurrence Theorem

### Centroid

### Coordinates of Centroid

### Identify Similar Triangles

### Prove Similar Triangles

### Identify Corresponding Sides

### Right Triangle Altitude Theorem (Part I)

### Right Triangle Altitude Theorem (Part II)

### Application Example 1

### Application Example 1

### Application Example 2

### Application Example 2

### Application Example 3

### Application Example 3

### Application Example 4

### Application Example 4

### Application Example 5

### Application Example 5

### Pythagorean Theorem

### Converse of Pythagorean Theorem

### Converse of Pythagorean Theorem

### Pythagorean Theorem

### Pythagorean Example

### Pythagorean Example

### Pythagorean Example

### Pythagorean Example

### Pythagorean Example

### Pythagorean Example

### Pythagorean Triples

### 45-45-Degree Right Triangle

### 30-60-Degree Right Triangle

### Special Right Triangle Example

### Special Right Triangle Example

### Special Right Triangle Example

### Special Right Triangle Example

### Triangle Angle Bisector Theorem

### Triangle Angle Bisector Theorem

### Triangle Angle Bisector Theorem

### Triangle Angle Bisector Theorem - Example 1

### Triangle Angle Bisector Theorem - Example 1

### Triangle Angle Bisector Theorem - Example 2

### Triangle Angle Bisector Theorem - Example 2

### Application Problem 1

### Application Problem 1

### Application Problem 2

### Application Problem 2

### Key Questions

### Steps of Proofs

### Example - In Proportions

### Example - Geometric Mean

### Example - Equal Product

### Triangle Proportionality Theorem(Side-Splitter Theorem)

### Triangle Angle Bisector Theorem

### Similar Triangles - AA

### Similar Triangles - SAS

### Similar Triangles - Parallel Sides & Shared Angle

### Similar Triangles - Parallel Sides & Vertical Angles

### Similar Triangles - Overlapping Triangles

### Similar Triangles - Angle Bisector

### Application Problem 1

### Application Problem 2

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

or 4 to 5 or 4 : 5

=

=

Mr. Chin-Sung Lin

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

or 4 to 1 or 4 : 1

=

=

Mr. Chin-Sung Lin

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,

?

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

200 m/s

=

=

Mr. Chin-Sung Lin

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 c e g

b d f h

=

or a : b = c : d

=

=

=

Mr. Chin-Sung Lin

A proportion a : b = c : d

a : b = c : d

extremes

means

Mr. Chin-Sung Lin

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

y

x

is the constant of proportionality

k =

Mr. Chin-Sung Lin

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

then a x d = b x c

=

Mr. Chin-Sung Lin

In a proportion, the extremes may be interchanged

a c d c

b d b a

b ≠ 0, d ≠ 0, a ≠ 0

then

=

=

Mr. Chin-Sung Lin

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

The proportions are equivalent when inverse both ratios

a c b d

b d a c

b ≠ 0, d ≠ 0, a ≠ 0, c ≠ 0

then

=

=

Mr. Chin-Sung Lin

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

a x b = c x d then

=

Mr. Chin-Sung Lin

The proportions are equivalent when adding unity to both ratios

a c a + b c + d

b d b d

b ≠ 0, d ≠ 0

then

=

=

Mr. Chin-Sung Lin

The proportions are equivalent when subtracting unity to both ratios

a c a - b c - d

b d b d

b ≠ 0, d ≠ 0

then

=

=

Mr. Chin-Sung Lin

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

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?

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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?

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

Solve the proportions for x

2 x + 2

5 2x - 1

=

Mr. Chin-Sung Lin

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

=

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

x (4x) = 82

4x2 = 64

x2 = 16

x = 4

Mr. Chin-Sung Lin

- 4x2 15y2
- 3 2
- 8x2 = 45 y2
- 4 3
- 15y2 2x2
- 5 4x
- 2x 9y2

If

, 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

- 4x2 15y2
- 3 2
- 8x2 = 45 y2
- 4 3
- 15y2 2x2
- 5 4x
- 2x 9y2

If

, 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

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

C

E

D

A

B

Mr. Chin-Sung Lin

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

C

E

F

D

A

B

Mr. Chin-Sung Lin

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

C

E

F

D

A

B

Mr. Chin-Sung Lin

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

E is the midpoint of BC

Prove: DE || AB, and

DE = ½ AB

C

E

F

D

A

B

Mr. Chin-Sung Lin

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

E is the midpoint of BC

Prove: DE || AB, and

DE = ½ AB

C

E

F

D

A

B

Mr. Chin-Sung Lin

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

C

E

D

A

B

Mr. Chin-Sung Lin

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

A

E

D

B

C

Mr. Chin-Sung Lin

D

E

B

C

ERHS Math Geometry

Statements Reasons

1. AD/DB = AE/EC1. Given

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

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

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

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

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

Mr. Chin-Sung Lin

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

A

E

D

B

C

Mr. Chin-Sung Lin

D

E

B

C

ERHS Math Geometry

Statements Reasons

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

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

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

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

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

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

Mr. Chin-Sung Lin

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

A

E

D

B

C

Mr. Chin-Sung Lin

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

A

E

D

B

C

Mr. Chin-Sung Lin

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

A

E

D

B

C

Mr. Chin-Sung Lin

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?

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

- 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)

Mr. Chin-Sung Lin

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

corresponding angles are congruent

6

6

4

60o

60o

6

12

9

corresponding sides are proportional

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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)

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

=

=

=

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

Mr. Chin-Sung Lin

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

=

=

=

y = 20

x = 10

z = 15

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

=

=

Mr. Chin-Sung Lin

Any geometric figure is similar to itself

ABC~ABC

Mr. Chin-Sung Lin

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

If ABC~DEF, then DEF~ABC

Mr. Chin-Sung Lin

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

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

X

R

Z

T

Y

S

C

B

ERHS Math Geometry

Statements Reasons

1. A X, and C Z1. Given

2. Draw RST ~ ABC 2. Postulate of similarity

with RT/AC = XZ/AC

3. R A, TC 3. Definition of similar triangles

4. R X, TZ 4. Transitive property

5. AC =AC 5. Reflecxive postulate

6. RT = XZ 6. Multiplication postulate

7. RST XYZ 7. SAS postulate

8. RST ~XYZ 8. Congruent ’s are similar ’s

9. ABC ~XYZ 9. Transitive property of similarity

A

C

45o

45o

D

B

ERHS Math Geometry

Given: mA = 45 and mD = 45

Prove: ABC~DEC

Mr. Chin-Sung Lin

A

45o

45o

C

D

B

ERHS Math Geometry

Statements Reasons

1. mA = 45 and mD = 45 1. Given

2. A D 2. Substitution property

3. ACB DCE 3. Vertical angles

4. ABC~DEC 4. AA similarity theorem

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

E

D

Mr. Chin-Sung Lin

X

Z

Y

C

B

ERHS Math Geometry

Statements Reasons

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

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

with AD = XY, DE || BC

3. ADE B, and AED C3. Corresponding angles

4. ADE ~ABC 4. AA similaity theorem

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

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

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

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

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

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

10. ADE XYZ 10. SSS postulate

11. ADE ~XYZ 11. Congruent ’s are similar ’s

12. ABC ~XYZ 12. Transitive property of similarity

E

D

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

Mr. Chin-Sung Lin

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

E

D

Mr. Chin-Sung Lin

X

Z

Y

C

B

ERHS Math Geometry

Statements Reasons

1. A X, AB/XY = AC/XZ1. Given

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

with AD = XY, DE || BC

3. ADE B, and AED C3. Corresponding angles

4. ADE ~ABC 4. AA similaity theorem

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

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

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

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

9. AE = XZ 9. Division postulate

10. ADE XYZ 10. SAS postulate

11. ADE ~XYZ 11. Congruent ’s are similar ’s

12. ABC ~XYZ 12. Transitive property of similarity

E

D

6

?

16

A

C

D

8

10

12

B

ERHS Math Geometry

Prove: ABC~DEC

Calculate: DE

Mr. Chin-Sung Lin

6

5

16

A

C

D

8

10

12

B

ERHS Math Geometry

Prove: ABC~DEC

Calculate: DE

Mr. Chin-Sung Lin

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

=

Mr. Chin-Sung Lin

D

E

B

C

ERHS Math Geometry

Statements Reasons

1. BC || DE1. Given

2. A A 2. Reflexive property

3. ABC ADE 3. Corresponding angles

4. ABC~ADE 4. AA similarity theorem

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

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

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

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

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

Mr. Chin-Sung Lin

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

=

=

=

Mr. Chin-Sung Lin

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

=

Mr. Chin-Sung Lin

D

E

B

C

ERHS Math Geometry

Statements Reasons

1. AD/DB = AE/EC1. Given

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

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

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

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

6. A A 6. Reflexive property

7. ABC ~ ADE 7. SAS similarity theorem

8. ABC ADE 8. Corresponding angles

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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)

Mr. Chin-Sung Lin

C (kc, 0)

B (kb, 0)

ERHS Math Geometry

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

y

y

X (0, a)

x

x

Z (c, 0)

Y (b,0)

Mr. Chin-Sung Lin

C (kc, 0)

B (kb, 0)

ERHS Math Geometry

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

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

C (kc, 0)

B (kb, 0)

ERHS Math Geometry

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

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

C (kc, 0)

B (kb, 0)

ERHS Math Geometry

Under a dilation, angle measure is preserved

y

y

X (0, a)

x

x

Z (c, 0)

Y (b,0)

Mr. Chin-Sung Lin

C (kc, 0)

B (kb, 0)

ERHS Math Geometry

Under a dilation, angle measure is preserved

We have proved:

y

y

X (0, a)

x

x

Z (c, 0)

Y (b,0)

X A

YB

ZC

Mr. Chin-Sung Lin

Under a dilation, midpoint is preserved

y

y

A (0, ka)

X (0, a)

N

M

x

x

Y (b, 0)

B (kb, 0)

Mr. Chin-Sung Lin

Under a dilation, midpoint is preserved

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

Under a dilation, collinearity is preserved

A (ka, kb)

y

Q (ke, kf)

X (a, b)

P (e, f)

B (kc, kd)

Y (c, d)

x

Mr. Chin-Sung Lin

Under a dilation, collinearity is preserved

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

Under a dilation, collinearity is preserved

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

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?

Mr. Chin-Sung Lin

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?

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

- If two triangles are similar, their corresponding
- sides
- altitudes
- medians, and
- angle bisectors

- are proportional

Mr. Chin-Sung Lin

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

X

Z

Y

Mr. Chin-Sung Lin

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

X

Z

Y

AA Similarity

Mr. Chin-Sung Lin

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

X

Z

Y

Mr. Chin-Sung Lin

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

X

Z

Y

SAS Similarity

Mr. Chin-Sung Lin

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

X

Z

Y

Mr. Chin-Sung Lin

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

X

Z

Y

AA Similarity

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

- 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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

A

C

ERHS Math Geometry

If BD is the median of ∆ ABC

then,

AD CD

D

Mr. Chin-Sung Lin

Centroid

A

C

ERHS Math Geometry

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

Mr. Chin-Sung Lin

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

B

E

D

Centroid

1

1

P

2

2

A

C

Mr. Chin-Sung Lin

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

B

E

D

1

Centroid

1

1

P

2

2

2

A

C

Mr. Chin-Sung Lin

Statements Reasons

1. AE and CD are medians 1. Given

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

E is the midpoint of BC

3. AC || DE3. Midsegment theorem

DE = ½ AC

4. AED EAC, CDE DCA 4. Alternate interior angles

5. APC ~ EPD 5. AA similarity theorem

6. AC:DE = 2:1 6. Exchange extremes

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

8. AP:EP = CP:DP = 2:1 8. Transitive property

B

E

D

1

1

1

P

2

2

2

A

C

Mr. Chin-Sung Lin

The medians of a triangle are concurrent

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

Prove: AM, BN, and CL are concurrent

B

L

M

P

Centroid

A

C

N

Mr. Chin-Sung Lin

- 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

B

L

M

P

Centroid

P’

A

C

N

Mr. Chin-Sung Lin

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

B

2

1

1

Centroid

2

2

1

A

C

Mr. Chin-Sung Lin

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

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

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

B

,

P (

)

P

Centroid

A

C

Mr. Chin-Sung Lin

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

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

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

D

C

B

ERHS Math Geometry

ABC ~ BDC ABC ~ ADB

AB AC BC AB AC BC

BD BC DC AD AB BD

BC2 = AC DC AB2 = AC AD

ADB ~ BDC

AD AB BD

BD BC CD

BD2 = AD DC

=

=

=

=

=

=

Mr. Chin-Sung Lin

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

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

2

D

y

8

x

C

z

B

ERHS Math Geometry

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

Solve: x, y, and z

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

6

D

12

x

y

C

z

B

ERHS Math Geometry

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

Solve: x, y, and z

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Solve: x, y, and z

A

z

D

6

9

y

C

x

B

Mr. Chin-Sung Lin

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

Solve: x, y, and z

x = 6√3

y = 3√3

z = 3

A

z

D

6

9

y

C

x

B

Mr. Chin-Sung Lin

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

Solve: w, x, y, and z

A

y

25

D

w

z

x

C

B

20

Mr. Chin-Sung Lin

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

Solve: w, x, y, and z

w = 15

x = 12

y = 9

z = 16

A

y

25

D

w

z

x

C

B

20

Mr. Chin-Sung Lin

Given: ABC, mB = 90

Prove: AB2 + BC2 = AC2 (Pythagorean Theorem)

(Hint: Apply Triangle Altitude Theorem)

A

D

C

B

Mr. Chin-Sung Lin

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

A

D

C

B

Mr. Chin-Sung Lin

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

c

A

b

a

B

C

Mr. Chin-Sung Lin

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

c

A

b

a

B

C

Mr. Chin-Sung Lin

c

b

b

Statements Reasons

1. a2 + b2 = c21. Given

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

m T = 90

3. ST2 + RT2 = RS2 3. Pythagorean theorem

4. a2 + b2 = RS2, RS2 = c2 4. Substitution postulate

5. RS = c5. Root postulate

6. ABC RST 6. SSS postulate

7. C T 7. CPCTC

8. mC = 90 8. Substitution postulate

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

a

a

A

R

B

S

C

T

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

c

A

b

a

B

C

Mr. Chin-Sung Lin

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?

A

15

?

B

C

D

24

Mr. Chin-Sung Lin

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

A

15

?

B

C

D

24

Mr. Chin-Sung Lin

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?

A

?

24

C

10

B

Mr. Chin-Sung Lin

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

A

?

24

C

10

B

Mr. Chin-Sung Lin

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?

A

?

16

12

B

15

Mr. Chin-Sung Lin

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

A

?

16

12

B

15

Mr. Chin-Sung Lin

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}

Mr. Chin-Sung Lin

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

45o

√ 2

1

A

45o

1

B

C

Mr. Chin-Sung Lin

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

A

60o

2

1

30o

B

C

√ 3

Mr. Chin-Sung Lin

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

4

?

A

B

C

Mr. Chin-Sung Lin

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

4

2√ 2

2√ 2

A

B

C

Mr. Chin-Sung Lin

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

A

4

?

30o

B

C

?

Mr. Chin-Sung Lin

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

A

4

2

30o

B

C

2√ 3

Mr. Chin-Sung Lin

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

=

Mr. Chin-Sung Lin

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

=

Mr. Chin-Sung Lin

A

1

2

3

C

B

D

ERHS Math Geometry

Statements Reasons

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

2. 1 2 2. Given

3. 2 3 3. 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/DC 7. Triangle proporatationality

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

Mr. Chin-Sung Lin

2

1

6

4

C

B

D

4

?

ERHS Math Geometry

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

Mr. Chin-Sung Lin

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

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

C

10

?

D

6

1

2

A

B

8

Mr. Chin-Sung Lin

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

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

Mr. Chin-Sung Lin

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

Mr. Chin-Sung Lin

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

Calculate BC

A

2

1

6

4

C

B

D

x

x - 1

Mr. Chin-Sung Lin

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

Calculate BC

x = 3

BC = 5

A

2

1

6

4

C

B

D

x

x - 1

Mr. Chin-Sung Lin

- Proving line segments are in proportion
- Proving line segments have geometric mean (mean proportional)
- Proving products of line segments are equal

Mr. Chin-Sung Lin

- 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

Mr. Chin-Sung Lin

A

C

D

B

ERHS Math Geometry

Given: ABC DEC

Prove: AC BC

CD CE

=

Mr. Chin-Sung Lin

Given: ABC BDC

Prove: BC is geometric mean between AC and DC

A

D

C

B

Mr. Chin-Sung Lin

Given: AB || DE

Prove: AD * CE = BE * DC

A

D

C

E

B

Mr. Chin-Sung Lin

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

=

=

=

Mr. Chin-Sung Lin

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

A

2

1

=

C

B

D

Mr. Chin-Sung Lin

X

Z

Y

C

B

ERHS Math Geometry

A X, B Y

ABC~XYZ

AB BC CA

XY YZ ZX

=

=

Mr. Chin-Sung Lin

X

Z

Y

C

B

ERHS Math Geometry

A X, AB/XY = AC/XZ

ABC~XYZ

AB BC

XY YZ

=

Mr. Chin-Sung Lin

C

B

ERHS Math Geometry

DE || BC

ABC~ADE

A A, B ADE, C AED

AB BC AC

AD DE AE

=

=

E

D

Mr. Chin-Sung Lin

DE || AB

ABC~DEC

A D, B E, ACB DCE

AB BC AC

DE EC DC

E

=

=

A

C

D

B

Mr. Chin-Sung Lin

D

C

B

ERHS Math Geometry

ABC BDC

ABC~BDC

ABC BDC, A DBC, C C

AB AC BC

BD BC DC

=

=

Mr. Chin-Sung Lin

1 2, (Draw EC || AD, 2 3)

ABD ~ EBC

B B, 1 E, ADB ECB

BA BD

AE DC

BA BD

AC DC

E

=

A

2

1

=

3

B

C

D

Mr. Chin-Sung Lin

Given: BC || DE, 1 2

Prove: EC AE

BC AC

=

A

E

D

1

2

C

B

Mr. Chin-Sung Lin

F

E

A

C

H

ERHS Math Geometry

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

Prove: AH * EF = BF * EH

Mr. Chin-Sung Lin

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