PROPORTIONAL PROPORTIONAL SEGMENTSSEGMENTS

&&BASIC SIMILARITY BASIC SIMILARITY

THEOREMTHEOREM

Express each ratio in simplest form.

18

12

12

8

14

8

ILLUSTRATIONILLUSTRATION

Pt. X divides Pt. X divides segment AB segment AB so that AX to so that AX to XB is 3 : 2.XB is 3 : 2.

Pt. Y divides Pt. Y divides segment CD segment CD so that CY to so that CY to YD is 3 : 2YD is 3 : 2

A

DYC

X B12 8

6 4

THEOREM: THEOREM: PROPORTIONAL SEGMENTSPROPORTIONAL SEGMENTS ““Two segments are divided proportionally Two segments are divided proportionally

if if the measures of the segments of one the measures of the segments of one have the same ratiohave the same ratio as the measures of as the measures of the corresponding segments of the other.”the corresponding segments of the other.”

A

DYC

X B12 8

6 4AX CY

12

XB YD

6 8 4

SOME PROPORTIONS

3 3

2 2

1.

THEOREM: THEOREM: PROPORTIONAL SEGMENTSPROPORTIONAL SEGMENTS ““Two segments are divided proportionally Two segments are divided proportionally

if if the measures of the segments of one the measures of the segments of one have the same ratiohave the same ratio as the measures of as the measures of the corresponding segments of the other.”the corresponding segments of the other.”

A

DYC

X B12 8

6 4AB CD

20

XB YD

108 4

SOME PROPORTIONS

5 5

2 2

2.

THEOREM: THEOREM: PROPORTIONAL SEGMENTSPROPORTIONAL SEGMENTS ““Two segments are divided proportionally Two segments are divided proportionally

if if the measures of the segments of one the measures of the segments of one have the same ratiohave the same ratio as the measures of as the measures of the corresponding segments of the other.”the corresponding segments of the other.”

A

DYC

X B12 8

6 4AB CD

20

AX CY

1012 6

SOME PROPORTIONS

5 5

3 3

3.

THEOREM: THEOREM: PROPORTIONAL SEGMENTSPROPORTIONAL SEGMENTS ““Two segments are divided proportionally Two segments are divided proportionally

if if the measures of the segments of one the measures of the segments of one have the same ratiohave the same ratio as the measures of as the measures of the corresponding segments of the other.”the corresponding segments of the other.”

A

DYC

X B12 8

6 4AX +AB CY +CD

32

AX CY

1612 6

SOME PROPORTIONS

8 8

3 3

4.

Illustrative Examples

Suppose segment AC and segment MP are Suppose segment AC and segment MP are divided proportionally by points B and N divided proportionally by points B and N respectively. Then,respectively. Then,

A

PNM

B C8 12

2 3AB MN

AB

BC NP

BCMN NP

1.

2.

Suppose segment AC and segment MP are Suppose segment AC and segment MP are divided proportionally by points B and N divided proportionally by points B and N respectively. Then,respectively. Then,

A

PNM

B C8 12

2 3AB MN

BC

AC MP

NPAC MP

3.

4.

Find the unknown parts assuming the Find the unknown parts assuming the segments are divided proportionally.segments are divided proportionally.

X 32

6 8

Solution:Solution:

X : 32 = 6 : 8X : 32 = 6 : 8

Applying the law of Applying the law of proportionproportion

8(x) = 6( 32)8(x) = 6( 32)

8x = 1928x = 192

X = 24X = 24

Illustrative examplesIllustrative examples

SOLVE:SOLVE: One string is divided into lengths 18 cm One string is divided into lengths 18 cm

and 15 cm. a second string is also to be and 15 cm. a second string is also to be divided into such that the two strings will divided into such that the two strings will become proportional. If the longest portion become proportional. If the longest portion of the second string has length 60 cm, of the second string has length 60 cm, what is the length of the other portion of what is the length of the other portion of the second string?the second string?

Solution:Solution: Let x = the length of the other Let x = the length of the other

portion of the second string.portion of the second string.

x 60

15 18

x18

60

6015

5x6

6x 5(60)

X = 50, the length of the other portion of the second string

6x = 300

BASIC PROPORTIONALITY BASIC PROPORTIONALITY THEOREMTHEOREM

If a line intersects If a line intersects two sides of a two sides of a triangle and is triangle and is parallel to the parallel to the third side, then it third side, then it divides the first divides the first two sides two sides proportionally.proportionally.

RESTATEMENT OF THE RESTATEMENT OF THE THEOREMTHEOREM

If a line (If a line (EFEF) ) intersects two sides intersects two sides ( ( AB & CBAB & CB) of a ) of a triangle (triangle (ABCABC) and is ) and is parallel to the third parallel to the third side(side( AC AC ), then it ), then it divides the first two divides the first two sides proportionally.sides proportionally.

Thus, Thus,

B

E F

CA

OTHER PROPORTIONSOTHER PROPORTIONS

1. BE : EA = BF : FC1. BE : EA = BF : FC 2. BE : BA = BF : BC2. BE : BA = BF : BC 3. BA : EA = BC : FC3. BA : EA = BC : FC 4. BE : BF = EA : FC4. BE : BF = EA : FC 5. FC : EA = BC : BA5. FC : EA = BC : BA 6. EF : AC = BF : BC6. EF : AC = BF : BC 7. EF : AC = BE : BA7. EF : AC = BE : BA

B

E F

CA

VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS( an example)( an example)

1. BE : EA = BF : 1. BE : EA = BF : FCFC

15 : 5 = 12 : 415 : 5 = 12 : 4 By simplifying,By simplifying, 3 : 1 = 3 : 13 : 1 = 3 : 1

B

E F

CA

15

4

12

58

6

VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS

2. BE : BA = BF : 2. BE : BA = BF : BCBC

15 : 20 = 12 : 15 : 20 = 12 : 1616

By simplifying,By simplifying,3 : 4 = 3 : 43 : 4 = 3 : 4

B

E F

CA

15

4

12

58

6

VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS

3. BA : EA = BC : 3. BA : EA = BC : FCFC

20 : 5 = 16 : 420 : 5 = 16 : 4

By simplifying,By simplifying,4 : 1 = 4 : 14 : 1 = 4 : 1

B

E F

CA

15

4

12

58

6

VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS

4. BE : BF = EA : 4. BE : BF = EA : FCFC

15 : 12 = 5 : 415 : 12 = 5 : 4 By simplifying,By simplifying,

5 : 4 = 5 : 45 : 4 = 5 : 4

B

E F

CA

15

4

12

58

6

VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS

5. FC : EA = BC : 5. FC : EA = BC : BABA

4 : 5 = 16 : 204 : 5 = 16 : 20 By simplifying,By simplifying,

4 : 5 = 4 : 54 : 5 = 4 : 5

B

E F

CA

15

4

12

58

6

VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS

6. EF : AC = BF : 6. EF : AC = BF : BCBC

6 : 8 = 12 : 166 : 8 = 12 : 16 By simplifying,By simplifying,

3 : 4 = 3 : 43 : 4 = 3 : 4

B

E F

CA

15

4

12

58

6

VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS

6. EF : AC = BE : 6. EF : AC = BE : BABA

6 : 8 = 15 : 206 : 8 = 15 : 20 By simplifying,By simplifying,

3 : 4 = 3 : 43 : 4 = 3 : 4

B

E F

CA

15

4

12

58

6

ExercisesExercises

GIVEN: DE // BC, GIVEN: DE // BC, AD = 9, AE = 12, AD = 9, AE = 12,

DE = 10,DB = 18.DE = 10,DB = 18.Find,Find,BC, AC and CE.BC, AC and CE.

A

D E

CB

9 12

18

10

Solution Solution Find BC,Find BC, BC : DE = BA : DABC : DE = BA : DA BC : 10 = 27 : 9 BC : 10 = 27 : 9 oror BC : 10 = 3 : 1BC : 10 = 3 : 1 Applying principle Applying principle

of proportionof proportion BC(1) = 10(3)BC(1) = 10(3) BC = 30BC = 30

A

D E

CB

9 12

18

10

30

Solution Solution Find AC,Find AC, AC : AE = BA : DAAC : AE = BA : DA AC : 12 = AC : 12 = 27 : 927 : 9 oror AC : 12 = 3 : 1AC : 12 = 3 : 1 Applying principle Applying principle

of proportionof proportion AC(1) = 12(3)AC(1) = 12(3)AC = 36AC = 36

A

D E

CB

9 12

18

10

Solution Solution Find CE,Find CE, CE : AE = BD : DACE : AE = BD : DA CE : 12 = 18 : 9CE : 12 = 18 : 9 oror CE : 12 = 2 : 1CE : 12 = 2 : 1 Applying principle Applying principle

of proportionof proportion CE(1) = 12(2)CE(1) = 12(2) CE = 24CE = 24

A

D E

CB

9 12

18

10

24

Solution Solution Another way to Another way to

find CE,find CE, CE = AC - AECE = AC - AE Hence, AC =36, Hence, AC =36,

thenthen CE = 36 - 12CE = 36 - 12 CE = 24CE = 24

A

D E

CB

9 12

18

10

24

Quiz Solve the following problem. Show your

solution.( one –half crosswise)

1. Uncle Tom plans to divide an 80- meter rope into three pieces in the ratio 3 : 5 : 8. what will be the length of each piece?

QUIZ

2. In the figure, find the values of x and y.

y15

10

30

12

x

Assignment EXERCISES A. Geometry Workbook, page 95