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Ratios, Proportions, and Similarity. Eleanor Roosevelt High School Chin-Sung Lin. ERHS Math Geometry. Ratio and Proportion. Mr. Chin-Sung Lin. ERHS Math Geometry. A ratio is a comparison by division of two quantities that have the same units of measurement - PowerPoint PPT Presentation
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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
Example - Geometric Mean
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
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
E is the midpoint of BC
Prove: DE || AB, and
DE = ½ AB
ERHS Math Geometry
C
A B
ED F
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 F
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 F
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 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
Divided Proportionally Theorem
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
Divided Proportionally Theorem
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
Application Examples
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
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
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
Example - Divided 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?
ERHS Math Geometry
Example - Divided 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
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
Definition of Similar Polygons
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
Definition of Similar Polygons
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
Example of Similar Triangles
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
Example of Similar Triangles
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
Example of Similar Triangles
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
= = =
Example of Similar Triangles
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
Example of Similar Triangles
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
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
DE
Side-Side-Side Similarity Theorem (SSS~)
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
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
DE
ERHS Math Geometry
Side-Angle-Side Similarity Theorem (SAS~)
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
Triangle Proportionality Theorem
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
Triangle Proportionality Theorem
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
Triangle Proportionality Theorem
Mr. Chin-Sung Lin
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 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
Triangle Proportionality Theorem
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
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
m XY = – a/bm XZ = – a/cm YZ = 0
m AB = – ka/kb = – a/bm AC = – ka/kc = – a/cm BC = 0
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
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)
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)
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)
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)
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
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
P2
1
Theorem about Centroid
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 || 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
Coordinates of Centroid
ERHS Math Geometry
Coordinates of Centroid
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
Application Examples
Mr. Chin-Sung Lin
ERHS Math Geometry
Application Example 1
Mr. Chin-Sung Lin
Given: ABC, mB = 90, BD is an altitude
Solve: x, y, and z
D
A
B C
8
2
yx
z
ERHS Math Geometry
Application Example 1
Mr. Chin-Sung Lin
Given: ABC, mB = 90, BD is an altitude
Solve: x, y, and z
x = 4
y = 2√5
z = 4√5
D
A
B C
8
2
yx
z
ERHS Math Geometry
Application Example 2
Mr. Chin-Sung Lin
Given: ABC, mB = 90, BD is an altitude
Solve: x, y, and z
D
A
B C
x
6
12y
z
ERHS Math Geometry
Application Example 2
Mr. Chin-Sung Lin
Given: ABC, mB = 90, BD is an altitude
Solve: x, y, and z
x = 18
y = 6√3
z = 12√3
D
A
B C
x
6
12y
z
ERHS Math Geometry
Application Example 3
Mr. Chin-Sung Lin
Given: ABC, mB = 90, BD is an altitude
Solve: x, y, and z
D
A
BC
z
6y
x
9
ERHS Math Geometry
Application Example 3
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
D
A
BC
z
6y
x
9
ERHS Math Geometry
Application Example 4
Mr. Chin-Sung Lin
Given: ABC, mB = 90, BD is an altitude
Solve: w, x, y, and z
D
A
B C
25y
wx
20
z
ERHS Math Geometry
Application Example 4
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
D
A
B C
25y
wx
20
z
ERHS Math Geometry
Application Example 5
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
Application Example 5
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?
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
42√ 2
2√ 2
Special Right Triangle Example
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
?
?
Special Right Triangle Example
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
Triangle Angle Bisector Theorem
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
Triangle Angle Bisector Theorem
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
E
3
ERHS Math Geometry
Triangle Angle Bisector Theorem
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
Triangle Angle Bisector Theorem - Example 1
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
Triangle Angle Bisector Theorem - Example 2
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
Triangle Angle Bisector Theorem - Example 2
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
Application Problem 1
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
Application Problem 2
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
Application Problem 2
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
Example - Geometric Mean
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
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
=
D E
A
B C
= =
ERHS Math Geometry
Triangle Angle Bisector Theorem
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
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
Application Problem 1
Mr. Chin-Sung Lin
Given: BC || DE, 1 2
Prove: EC AE
BC AC
DE
A
B C
12
=
ERHS Math Geometry
Application Problem 2
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