ratios, proportions, and similarity

Ratios, Proportions, and Similarity

Eleanor Roosevelt High School

Chin-Sung Lin

Ratio and Proportion

Mr. Chin-Sung Lin

ERHS Math Geometry

Definition of Ratio

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Definition of Ratio

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

ERHS Math Geometry

Definition of Ratio

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,

?

ERHS Math Geometry

Definition of Ratio

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

ERHS Math Geometry

Ratio Example

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

ERHS Math Geometry

Ratio Example

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

ERHS Math Geometry

Definition of Rate

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

ERHS Math Geometry

Definition of Proportion

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

== =

ERHS Math Geometry

Definition of Extremes and Means

Mr. Chin-Sung Lin

A proportion a : b = c : d

a : b = c : d

ERHS Math Geometry

extremes

means

Definition of Constant of Proportionality

Mr. Chin-Sung Lin

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

y xk = is the constant of proportionality

ERHS Math Geometry

Properties ofProportions

Mr. Chin-Sung Lin

ERHS Math Geometry

Cross-Product Property

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

ERHS Math Geometry

Interchange Extremes Property

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

ERHS Math Geometry

=

Interchange Means Property(Alternation Property)

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

ERHS Math Geometry

=

Inversion Property

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 =

ERHS Math Geometry

Equal Factor Products Property

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

=

ERHS Math Geometry

a x b = c x d then

Composition Property

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 =

ERHS Math Geometry

Division Property

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 =

ERHS Math Geometry

Definition of Geometric Mean (Mean Proportional)

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

ERHS Math Geometry

Application Examples

Mr. Chin-Sung Lin

ERHS Math Geometry

Example - Ratios and Rates

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?

ERHS Math Geometry

Example - Ratios and Rates

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

ERHS Math Geometry

Example - Ratios

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?

ERHS Math Geometry

Example - Ratios

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

ERHS Math Geometry

Example - Ratios

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

ERHS Math Geometry

Example - Ratios

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

ERHS Math Geometry

Example - Proportions

Mr. Chin-Sung Lin

Solve the proportions for x

2 x + 2

5 2x - 1 =

ERHS Math Geometry

Example - Proportions

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

=

ERHS Math Geometry

Example - Geometric Mean

Mr. Chin-Sung Lin

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

ERHS Math Geometry


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

ERHS Math Geometry

Example - Properties of Proportions

Mr. Chin-Sung Lin

4x2 15y2

3 2

(a) 8x2 = 45 y2

(b) 4 3

15y2 2x2

(a) 5 4x

2x 9y2

=If

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

, which of the following

=

=

=(d) 4x + 9y 5y +

2x

9y 2x

(e) 9y 2x

4x - 9y 5y - 2x

=

ERHS Math Geometry

Midsegment Theorem

Mr. Chin-Sung Lin

ERHS Math Geometry

Midsegment Theorem

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

ERHS Math Geometry

C

A B

ED

Midsegment Theorem

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


Prove: DE || AB, and

DE = ½ AB

ERHS Math Geometry

C

A B

ED F

Divided Proportionally Theorem

Mr. Chin-Sung Lin

ERHS Math Geometry

Definition of Divided Proportionally

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

ERHS Math Geometry

C

A B

ED


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

ERHS Math Geometry

A

B C

ED


Mr. Chin-Sung Lin

Statements Reasons

1. AD/DB = AE/EC 1. 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

D E

A

B C

ERHS Math Geometry

Converse of Divided Proportionally Theorem

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

ERHS Math Geometry

A

B C

ED

Converse of Divided Proportionally Theorem

Mr. Chin-Sung Lin

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

D E

A

B C

ERHS Math Geometry

Divided Proportionally Theorem & Converse of Divided Proportionally Theorem

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

ERHS Math Geometry

A

B C

ED


Mr. Chin-Sung Lin

ERHS Math Geometry

Example - Divided Proportionally

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

ERHS Math Geometry

A

B C

ED


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

ERHS Math Geometry

A

B C

ED


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?

ERHS Math Geometry


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

ERHS Math Geometry

Similar Triangles

Mr. Chin-Sung Lin

ERHS Math Geometry

Definition of Similar Figures

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Definition of Similar Polygons

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)

ERHS Math Geometry


Mr. Chin-Sung Lin

Both conditions must be true for polygons to be similar:

(1) Corresponding angles are congruent

(2) The lengths of corresponding sides are proportional

ERHS Math Geometry

A B

CD

4

6

P Q

S R

6

9

60o

W X

Z Y

6

1260o

corresponding angles are congruent

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

ERHS Math Geometry

Definition of Similar Triangles

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)

ERHS Math Geometry

Example of Similar Triangles

Mr. Chin-Sung Lin

A X, B Y, C Z

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

XY = 3, YZ = 4 and ZX = 5

Show that ABC~XYZX

Y Z

3

4

5

A

B C

6

8

10

ERHS Math Geometry


Mr. Chin-Sung Lin

A X, B Y, C Z

AB BC CA 2

XY YZ ZX 1

Therefore ABC~XYZ

= = =

ERHS Math Geometry

X

Y Z

3

4

5

A

B C

6

8

10


Mr. Chin-Sung Lin

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

4

6

8 ?

?

?

ERHS Math Geometry


Mr. Chin-Sung Lin

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

4

6

8 x = 10

z = 15

y = 20

ERHS Math Geometry

= = =


Mr. Chin-Sung Lin

In ABC, AB = 9, BC = 15, AC = 18.

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

ERHS Math Geometry

X

Y Z

?

?

129

15

18

A

B C


Mr. Chin-Sung Lin

Since ABC~XYZ, and XZ = 12, then

XY YZ 12

9 15 18

X

Y Z

6

10

129

15

18

A

B C

ERHS Math Geometry

= =

Equivalence Relation of Similarity

Mr. Chin-Sung Lin

ERHS Math Geometry

Reflexive Property

Mr. Chin-Sung Lin

Any geometric figure is similar to itself

ABC~ABC

ERHS Math Geometry

Symmetric Property

Mr. Chin-Sung Lin

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

If ABC~DEF, then DEF~ABC

ERHS Math Geometry

Transitive Property

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

ERHS Math Geometry

Postulates & Theorems

Mr. Chin-Sung Lin

ERHS Math Geometry

Postulate of Similarity

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Angle-Angle Similarity Theorem (AA~)

Mr. Chin-Sung Lin

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

X

Y Z

A

B C

ERHS Math Geometry

Angle-Angle Similarity Theorem (AA~)

Statements Reasons

1. A X, and C Z 1. Given

2. Draw RST ~ ABC 2. Postulate of similarity

with RT/AC = XZ/AC

3. R A, T C 3. Definition of similar triangles

4. R X, T Z 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 ’s9. ABC ~ XYZ 9. Transitive property of similarity

ERHS Math Geometry

X

Y Z

A

B C

R

S T

Example of AA Similarity Theorem

Mr. Chin-Sung Lin

Given: mA = 45 and mD = 45

Prove: ABC~DEC

45o

A

B

C 45o

D

E

ERHS Math Geometry

Example of AA Similarity Theorem

Mr. Chin-Sung Lin

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

45oA

B

C45o

D

E

ERHS Math Geometry

Side-Side-Side Similarity Theorem (SSS~)

Mr. Chin-Sung Lin

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

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

Prove: ABC~XYZ

ERHS Math Geometry

X

Y Z

A

B C


Mr. Chin-Sung Lin

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

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

Prove: ABC~XYZ

ERHS Math Geometry

X

Y Z

A

B C

DE


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 || BC3. ADE B, and AED C 3. Corresponding angles 4. ADE ~ ABC 4. AA similaity theorem5. AB/AD = AC/AE = BC/DE 5. Corresponding sides

proportional6. AB/XY = AC/AE = BC/DE 6. Substitutin postulate7. AC/AE = AC/XZ, BC/DE = BC/YZ 7. Transitive postulate8. (AC)(XZ) = (AC)(AE) 8. Cross product (BC)(DE) = (BC)(YZ)9. AE = XZ, DE = YZ 9. Division postulate10. ADE XYZ 10. SSS postulate11. ADE ~ XYZ 11. Congruent ’s are similar ’s12. ABC ~ XYZ 12. Transitive property of similarity

ERHS Math Geometry

X

Y Z

A

B C

DE

Side-Angle-Side Similarity Theorem (SAS~)

Mr. Chin-Sung Lin

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

X

Y Z

A

B C

ERHS Math Geometry


Mr. Chin-Sung Lin

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

X

Y Z

A

B C

DE

ERHS Math Geometry


Statements Reasons

1. A X, AB/XY = AC/XZ 1. Given 2. Draw DE, D is on AB, E is on AC 2. Postulate of similarity with AD = XY, DE || BC3. ADE B, and AED C 3. Corresponding angles 4. ADE ~ ABC 4. AA similaity theorem5. AB/AD = AC/AE 5. Corresponding sides proportional6. AB/XY = AC/AE 6. Substitutin postulate7. AC/AE = AC/XZ 7. Transitive postulate8. (AC)(XZ) = (AC)(AE) 8. Cross product9. AE = XZ 9. Division postulate10. ADE XYZ 10. SAS postulate11. ADE ~ XYZ 11. Congruent ’s are similar ’s12. ABC ~ XYZ 12. Transitive property of similarity

ERHS Math Geometry

X

Y Z

A

B C

DE

Example of SAS Similarity Theorem

Mr. Chin-Sung Lin

Prove: ABC~DEC

Calculate: DE

16A

B

CD

E

1012

8

6 ?

ERHS Math Geometry

Example of SAS Similarity Theorem

Mr. Chin-Sung Lin

Prove: ABC~DEC

Calculate: DE

16A

B

CD

E

1012

8

6 5

ERHS Math Geometry

Triangle Proportionality Theorem

(Side-Splitter Theorem)

Mr. Chin-Sung Lin

ERHS Math Geometry


Mr. Chin-Sung Lin

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 =

D E

A

B C

ERHS Math Geometry


Mr. Chin-Sung Lin

Statements Reasons

1. BC || DE 1. 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

D E

A

B C

ERHS Math Geometry


Mr. Chin-Sung Lin


DE || BC

AD AE

DB EC

AD AE DE

AB AC BCD E

A

B C

= =

ERHS Math Geometry

=

Converse of Triangle Proportionality Theorem

Mr. Chin-Sung Lin

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

=

D E

A

B C

ERHS Math Geometry

Converse of Triangle Proportionality Theorem

Mr. Chin-Sung Lin

Statements Reasons

1. AD/DB = AE/EC 1. 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

D E

A

B C

ERHS Math Geometry


Mr. Chin-Sung Lin

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

D E

A

B C

ERHS Math Geometry

Example of Triangle Proportionality Theorem

Mr. Chin-Sung Lin

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

Calculate: CE and BC

8

3

4 6

?D E

A

BC

?

ERHS Math Geometry

Example of Triangle Proportionality Theorem

Mr. Chin-Sung Lin

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

Calculate: CE and BC

8

3

4 6

4.5D E

A

BC

14

ERHS Math Geometry

Dilations

Mr. Chin-Sung Lin

ERHS Math Geometry

Dilations

Mr. Chin-Sung Lin

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

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

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

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

ERHS Math Geometry

Dilations

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)

ERHS Math Geometry

Dilations and Similar Triangles

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Y (b,0) Z (c, 0)

X (0, a)

x

yA (0, ka)

B (kb, 0) C (kc, 0)x

y


Mr. Chin-Sung Lin


ERHS Math Geometry

Y (b,0) Z (c, 0)

X (0, a)

x

yA (0, ka)

B (kb, 0) C (kc, 0)x

y

m XY = – a/bm XZ = – a/cm YZ = 0

m AB = – ka/kb = – a/bm AC = – ka/kc = – a/cm BC = 0


Mr. Chin-Sung Lin


ERHS Math Geometry

Y (b,0) Z (c, 0)

X (0, a)

x

yA (0, ka)

B (kb, 0) C (kc, 0)x

y

XY || ABXZ || ACYZ || BC

X AY BZ C

ABC ~ ADE

Dilations and Angle Measures

Mr. Chin-Sung Lin

Under a dilation, angle measure is preserved

ERHS Math Geometry

Y (b,0) Z (c, 0)

X (0, a)

x

yA (0, ka)

B (kb, 0) C (kc, 0)x

y

Dilations and Angle Measures

Mr. Chin-Sung Lin

Under a dilation, angle measure is preserved

We have proved:

ERHS Math Geometry

Y (b,0) Z (c, 0)

X (0, a)

x

yA (0, ka)

B (kb, 0) C (kc, 0)x

y

X AY BZ C

Dilations and Midpoint

Mr. Chin-Sung Lin

Under a dilation, midpoint is preserved

ERHS Math Geometry

M

Y (b, 0)

X (0, a)

x

y

A (0, ka)

B (kb, 0)x

y

N

Dilations and Midpoint

Mr. Chin-Sung Lin

Under a dilation, midpoint is preserved

ERHS Math Geometry

M

Y (b, 0)

X (0, a)

x

y

A (0, ka)

B (kb, 0)x

y

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

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

N

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

N(A, B) = DK (M)

Dilations and Collinearity

Mr. Chin-Sung Lin

Under a dilation, collinearity is preserved

ERHS Math Geometry

P (e, f)

X (a, b)

x

y A (ka, kb)

Y (c, d) B (kc, kd)

Q (ke, kf)


Mr. Chin-Sung Lin


ERHS Math Geometry

P (e, f)

X (a, b)

x

y A (ka, kb)

Y (c, d) B (kc, kd)

Q (ke, kf)

m XP = (b – f) / (a – e)m PY = (f – d) / (e – c)XPY are collinearm XP = m PY

(b – f) / (a – e) = (f – d) / (e – c)


Mr. Chin-Sung Lin


ERHS Math Geometry

P (e, f)

X (a, b)

x

y A (ka, kb)

Y (c, d) B (kc, kd)

Q (ke, kf)

m XP = (b – f) / (a – e)m PY = (f – d) / (e – c)XPY are collinearm XP = m PY

(b – f) / (a – e) = (f – d) / (e – c)

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

Dilations Example

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?

ERHS Math Geometry

Dilations Example

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?

ERHS Math Geometry

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)

Proportional Relations among Segments Related to Triangles

Mr. Chin-Sung Lin

ERHS Math Geometry

Proportional Relations of Segments

Mr. Chin-Sung Lin

If two triangles are similar, their corresponding

• sides

• altitudes

• medians, and

• angle bisectors

are proportional

ERHS Math Geometry

Proportional Altitudes

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Y Z

XA

B C

Proportional Altitudes

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Y Z

XA

B C

AA Similarity

Proportional Medians

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Y Z

XA

B C

Proportional Medians

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Y Z

XA

B C

SAS Similarity

Proportional Angle Bisectors

Mr. Chin-Sung Lin

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

ERHS Math Geometry

Y Z

XA

B C

Proportional Angle Bisectors

Mr. Chin-Sung Lin

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

ERHS Math Geometry

AA Similarity

Y Z

XA

B C

Application Problem

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

ERHS Math Geometry

Application Problem

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

ERHS Math Geometry

Concurrence of the Medians of a Triangle

Mr. Chin-Sung Lin

ERHS Math Geometry

Median of a Triangle

ERHS Math Geometry

Mr. Chin-Sung Lin

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

A

C

B

A C

B

C

A

B

Median of a Triangle

Mr. Chin-Sung Lin

If BD is the median of ∆ ABC

then,

AD CD

A C

B

D

ERHS Math Geometry

Definition of Centroid

ERHS Math Geometry

Mr. Chin-Sung Lin

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

A

B

Centroid

C

Theorem about Centroid

ERHS Math Geometry

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

A

B

Centroid

C

2

1 1

2

D E

P


ERHS Math Geometry

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

A

B

Centroid

C

2

1 1

2

D E

P2

1


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


3. AC || DE 3. 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

ERHS Math Geometry

A

B

C

2

1 1

2

D E

P2

1

Medians Concurrence Theorem

ERHS Math Geometry

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

A

B

Centroid

C

M

N

LP

Medians Concurrence Theorem

ERHS Math Geometry

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

A

B

Centroid

C

M

N

LP

P’

Centroid

ERHS Math Geometry

Mr. Chin-Sung Lin

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

A

B

Centroid

C

2

1

1 1

22

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

ERHS Math Geometry

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


ERHS Math Geometry

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

A

B

Centroid

C

PP ( ),

Proportions in a Right Triangle

Mr. Chin-Sung Lin

ERHS Math Geometry

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

Mr. Chin-Sung Lin

Projection of a Point on a Line

ERHS Math Geometry

P

ProjectionA

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

Mr. Chin-Sung Lin

Projection of a Segment on a Line

ERHS Math Geometry

P

ProjectionA

Q

B

Similar Triangles within a Right Triangle

Mr. Chin-Sung Lin

ERHS Math Geometry

Identify Similar Triangles

Mr. Chin-Sung Lin

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

A

B C

D

ERHS Math Geometry

Prove Similar Triangles

Mr. Chin-Sung Lin

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

A

B C

D

ERHS Math Geometry

Identify Corresponding Sides

Mr. Chin-Sung Lin

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

A

B C

D

= = = =

= =

ERHS Math Geometry

Right Triangle Altitude Theorem

Mr. Chin-Sung Lin

ERHS Math Geometry

Right Triangle Altitude Theorem (Part I)

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

B CD

ERHS Math Geometry

Right Triangle Altitude Theorem (Part II)

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

ERHS Math Geometry

A

B CD


Mr. Chin-Sung Lin

ERHS Math Geometry

Application Example 1

Mr. Chin-Sung Lin


Solve: x, y, and z

D

A

B C

8

2

yx

z

ERHS Math Geometry


Mr. Chin-Sung Lin


Solve: x, y, and z

x = 4

y = 2√5

z = 4√5

D

A

B C

8

2

yx

z

ERHS Math Geometry


Mr. Chin-Sung Lin


Solve: x, y, and z

D

A

B C

x

6

12y

z

ERHS Math Geometry


Mr. Chin-Sung Lin


Solve: x, y, and z

x = 18

y = 6√3

z = 12√3

D

A

B C

x

6

12y

z

ERHS Math Geometry


Mr. Chin-Sung Lin


Solve: x, y, and z

D

A

BC

z

6y

x

9

ERHS Math Geometry


Mr. Chin-Sung Lin


Solve: x, y, and z

x = 6√3

y = 3√3

z = 3

D

A

BC

z

6y

x

9

ERHS Math Geometry


Mr. Chin-Sung Lin


Solve: w, x, y, and z

D

A

B C

25y

wx

20

z

ERHS Math Geometry


Mr. Chin-Sung Lin


Solve: w, x, y, and z

w = 15

x = 12

y = 9

z = 16

D

A

B C

25y

wx

20

z

ERHS Math Geometry


Mr. Chin-Sung Lin

Given: ABC, mB = 90

Prove: AB2 + BC2 = AC2 (Pythagorean Theorem)

(Hint: Apply Triangle Altitude Theorem)

D

A

B C

ERHS Math Geometry


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

D

A

B C

ERHS Math Geometry

Pythagorean Theorem

Mr. Chin-Sung Lin

ERHS Math Geometry

Pythagorean Theorem

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

ERHS Math Geometry

A

C Ba

bc

Converse of Pythagorean Theorem

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

ERHS Math Geometry

A

C Ba

bc

Converse of Pythagorean Theorem

Statements Reasons

1. a2 + b2 = c2 1. 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 = c 5. 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

ERHS Math Geometry

A

C Ba

bc

R

T Sa

b

Pythagorean Theorem

Mr. Chin-Sung Lin

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

ERHS Math Geometry

A

C Ba

bc

Pythagorean Example

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?

ERHS Math Geometry

A

CB

15

D

24

?

Pythagorean Example

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

ERHS Math Geometry

A

CB

15

D

24

?

Pythagorean Example

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?

ERHS Math Geometry

A

C

24?

B 10

Pythagorean Example

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

ERHS Math Geometry

A

C

24?

B 10

Pythagorean Example

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?

ERHS Math Geometry

12

15

16

A

B

?

Pythagorean Example

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

ERHS Math Geometry

12

15

16

A

B

?

Pythagorean Triples

Mr. Chin-Sung Lin

ERHS Math Geometry

Pythagorean Triples

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}

ERHS Math Geometry

Special Right Triangles

Mr. Chin-Sung Lin

ERHS Math Geometry

45-45-Degree Right Triangle

Mr. Chin-Sung Lin

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

ERHS Math Geometry

A

CB

145o

45o

1√ 2

30-60-Degree Right Triangle

Mr. Chin-Sung Lin

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

ERHS Math Geometry

A

CB

2

30o

60o

1

√ 3

Special Right Triangle Example

Mr. Chin-Sung Lin

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

ERHS Math Geometry

A

C B

4?


Mr. Chin-Sung Lin

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

ERHS Math Geometry

A

C B

42√ 2

2√ 2


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?

ERHS Math Geometry

A

C B

4

30o

?

?


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?

ERHS Math Geometry

A

C B

4

30o

2

2√ 3

Triangle Angle Bisector Theorem

Mr. Chin-Sung Lin

ERHS Math Geometry


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 =

D

1

A

B C

2

ERHS Math Geometry


Mr. Chin-Sung Lin


1 2

AB BD

AC DC =

D

1

A

B C

2

E

3

ERHS Math Geometry


Mr. Chin-Sung Lin

Statements Reasons

1. Extend BA and draw EC || AD 1. 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

D

1

A

B C

2

E

3

ERHS Math Geometry

Triangle Angle Bisector Theorem - Example 1

Mr. Chin-Sung Lin

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

D

1

A

B C

26

4

4 ?

ERHS Math Geometry


Mr. Chin-Sung Lin

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

AB BD

AC DC

6 4

4 DC

DC = 8/3D

1

A

B C

26

4

4 8/3

=

=

ERHS Math Geometry


Mr. Chin-Sung Lin

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

D

1

AB

C

2

6

8

10

?

ERHS Math Geometry


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

=

=D

1

AB

C

2

6

8

10

40/3

ERHS Math Geometry

Application Problems

Mr. Chin-Sung Lin

ERHS Math Geometry

Application Problem 1

Mr. Chin-Sung Lin

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

Calculate AB

D

1

A

B C

23x - 2

6

x + 1 3

ERHS Math Geometry


Mr. Chin-Sung Lin

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

Calculate AB

x = 4

AB = 10

D

1

A

B C

23x - 2

6

x + 1 3

ERHS Math Geometry


Mr. Chin-Sung Lin

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

Calculate BC

D

1

A

B C

26

4

x x - 1

ERHS Math Geometry


Mr. Chin-Sung Lin

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

Calculate BC

x = 3

BC = 5

D

1

A

B C

26

4

x x - 1

ERHS Math Geometry

Proportions & Products of Line Segments

Mr. Chin-Sung Lin

ERHS Math Geometry

Key Questions

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

ERHS Math Geometry

Steps of Proofs

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

ERHS Math Geometry

Example - In Proportions

Mr. Chin-Sung Lin

Given: ABC DEC

Prove: AC BC

CD CE

A

B

CD

E

=

ERHS Math Geometry


Mr. Chin-Sung Lin

Given: ABC BDC

Prove: BC is geometric mean between AC and DC

A

B

CD

ERHS Math Geometry

Example - Equal Product

Mr. Chin-Sung Lin

Given: AB || DE

Prove: AD * CE = BE * DC

A

B

CD

E

ERHS Math Geometry

Proportionality TheoremReview

Mr. Chin-Sung Lin

ERHS Math Geometry

Triangle Proportionality Theorem(Side-Splitter Theorem)

Mr. Chin-Sung Lin


DE || BC

AD AE

DB EC

AD AE DE

AB AC BC

=

D E

A

B C

= =

ERHS Math Geometry


Mr. Chin-Sung Lin


1 2

AB BD

AC DC =

D

1

A

B C

2

ERHS Math Geometry

Similar Triangles Review

Mr. Chin-Sung Lin

ERHS Math Geometry

Similar Triangles - AA

Mr. Chin-Sung Lin

A X, B Y

ABC~XYZ

AB BC CA

XY YZ ZX

X

Y Z

A

B C

= =

ERHS Math Geometry

Similar Triangles - SAS

Mr. Chin-Sung Lin

A X, AB/XY = AC/XZ

ABC~XYZ

AB BC

XY YZ

X

Y Z

A

B C

=

ERHS Math Geometry

Similar Triangles - Parallel Sides & Shared Angle

Mr. Chin-Sung Lin

DE || BC

ABC~ADE

A A, B ADE, C AED

AB BC AC

AD DE AE

DE

A

B C

= =

ERHS Math Geometry

Similar Triangles - Parallel Sides & Vertical Angles

Mr. Chin-Sung Lin

DE || AB

ABC~DEC

A D, B E, ACB DCE

AB BC AC

DE EC DC = =

A

B

C

D

E

ERHS Math Geometry

Similar Triangles - Overlapping Triangles

Mr. Chin-Sung Lin

ABC BDC

ABC~BDC

ABC BDC, A DBC, C C

AB AC BC

BD BC DCA

B

CD

= =

ERHS Math Geometry

Similar Triangles - Angle Bisector

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

=

=

D

1

A

B C

2

E

3

ERHS Math Geometry

Application Problems

Mr. Chin-Sung Lin

ERHS Math Geometry


Mr. Chin-Sung Lin

Given: BC || DE, 1 2

Prove: EC AE

BC AC

DE

A

B C

12

=

ERHS Math Geometry


Mr. Chin-Sung Lin

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

Prove: AH * EF = BF * EH

F

E

A

B

CH

ERHS Math Geometry

Q & A

Mr. Chin-Sung Lin

ERHS Math Geometry

The End

Mr. Chin-Sung Lin

ERHS Math Geometry

ratios, proportions, and similarity

Documents