CONFIDENTIAL 1

Geometry

Properties of Parallelogram

CONFIDENTIAL 2

Warm up

1) 120°

2) 135°

3) 156°

An interior angle measure of a regular polygon is given. Find the number of sides and the

measure of each exterior angle:

CONFIDENTIAL 3

Any polygon with four sides is a quadrilateral. However, some quadrilaterals have special properties. These special

quadrilaterals have their own names.

A quadrilaterals with two pairs of parallel sides is a parallelogram. To write the name of a parallelogram, you

use the symbol □.

Properties of Parallelograms

parallelogram ABCD

□ ABCD

A B

CD

AB || CD; BC || DA

CONFIDENTIAL 4

Theorem Hypothesis Conclusion

If a quadrilateral is a parallelogram, then its opposite sides are congruent. (□ -> opp. sides )

Properties of Parallelograms

A B

CD

AB CDBC DA

Theorem 1:

CONFIDENTIAL 5

Proof of Theorem 1:A B

CD

1

2

3

4

Given: ABCD is a parallelogram.Proof: AB CD, BC DA

Proof:

STATEMENTS REASONS

1. ABCD is a parallelogram 1. Given2. AB CD, BC DA 2. Def. of gm3. 1 2, 3 4 3. Alt. int. s Thm.4. AC AC 4. Reflex. prop. of .5. ABC CDA 5. ASA Steps 3,46. AB CD, BC DA 6. CPCTC

CONFIDENTIAL 6

Theorem Hypothesis Conclusion

If a quadrilateral is a parallelogram, then its opposite angles are congruent. (□ -> opp. angles )

Properties of Parallelograms

A B

CD

A CB D

CONFIDENTIAL 7

Theorem Hypothesis Conclusion

If a quadrilateral is a parallelogram, then its consecutive angles are supplementary. (□ -> cons. ∕s supp.)

Properties of Parallelograms

A B

CD

mA + B 180mB + C 180mC + D 180mD + A 180

CONFIDENTIAL 8

Theorem Hypothesis Conclusion

If a quadrilateral is a parallelogram, then its diagonals bisect each other.(□ -> diags. bisect each other)

Properties of Parallelograms

A B

CD

1

2

3

4

ZAZ CZBZ DZ

CONFIDENTIAL 9

In □ PQRS, QR = 48 cm, RT = 30 cm, and /QPS = 73°. Find each measure.

Racing application

Q R

SP

TA) PS PS QR gm -> opp. angles PS QR Def. of segs. PS = 48 cm Substitute 48 for QR

B) mPQR mPQR + mQPS = 180 gm -> opp. angles mPQR + 73 = 180 Substitute 73 for mQPS mPQR = 107 Substract 73 from both sides

C) PT PT RT gm -> diag. bisect each other PT RT Def. of segs. PT = 48 cm Substitute 30 for RT

CONFIDENTIAL 10

Now you try!

In □KLMN, LM = 28 in., LN = 26 in. and m/LKN = 74°. Find each measure:

1a) KN1b) m/NML1c) LO

M L

KN

O

CONFIDENTIAL 11

ABCD is a parallelogram. Find each measure.

Using Properties of Parallelogram to Find Measures

B C

DA

5x+19

(6y+5)°

(10y-1)°

A) AD AD BC gm -> opp. sides AD BC Def. of segs. 7x = 5x+19 Substitute the given values 2x = 19 Subtract 5x from both sides x = 9.5 Divide both sides by 2 AD = 7x = 7(9.5) = 66.5

CONFIDENTIAL 12

B C

DA

5x+19

(6y+5)°

(10y-1)°

B) mB mA + mB = 180 gm -> opp. angles (10y-1) + (6y+5) = 180 Substitute the given values 16y + 4 = 180 Combine like terms 16y = 176 Substract 4 from both sides y = 11 Divide both sides by 16 B =(6y+5) = [6(11)+5] = 71

CONFIDENTIAL 13

Now you try!

ABCD is a parallelogram. Find each measure:

E

F G

H

J

4z-9

2z

w+8

3w2a) JG2b) FH

CONFIDENTIAL 14

Three vertices of □ABCD are A(1,-2), B(-2,3) and D(5,-1). Find the coordinates of vertex C.

Parallelograms in the coordinate plane

B

A

D

5

60

C

-3

-2

-3

5

x

y

Since ABCD is a parallelogram, both pairs of opposite sides must

be parallel.

Step1: Graph the given points.

Step2: Find the slope of AB by counting the units from A to B.The rise from -2 to 3 is 5.The run from 1 to -2 is -3.

CONFIDENTIAL 15

B

A

D

5

60

C

-3

-2

-3

5

x

y

Step3: Start at D and count the same number of units.

The rise from -1 to 4 is 5.The run from 5 to 2 is -3.Label (2,4) as vertex C.

Step4: Use the slope formula to verify that BC || AD.

Slope of BC = 4 – 3 = 1 2 – (-2) 4

Slope of AD = -1 – (-2) = 1 5 – 1 4

The coordinates of vertex C are (2, 4).

CONFIDENTIAL 16

Now you try!

3) Three vertices of □PQRS are P(-3,-2), Q(-1,4) and S(5,0). Find the coordinates of vertex R.

CONFIDENTIAL 17

Using properties of Parallelograms in a proof

Given: ABCD is a parallelogram.Proof: BAD DCB, ABC CDA

Proof:

STATEMENTS REASONS

1. ABCD is a parallelogram 1. Given2. AB CD, BC DA 2. gm opp.sides 3. BD BD 3. Reflex. prop. of .4. BAF DCB 4. SSS Steps 2,35. BAD DCB 5. CPCTC6. AC AC 6. Reflex. prop. of .7. ABC CDA 7. SSS Steps 2,68. ABC CDA 8. CPCTC

B C

DA

E

A)

CONFIDENTIAL 18

B) Given: GHJN and JKLM are paralellograms. H and M are collinear.Prove: G L

Proof:

STATEMENTS REASONS

1. GHJN and JKLM are grams 1. Given2.HJN G, MJK L 2. gm opp. s 3. HJN MJK 3. Vert. s Thm.4. G L 4. Trans. prop. of .

J

H

G

N M

L

K

CONFIDENTIAL 19

Now you try!

4) Use the figure above to write a two-column proof.

J

H

G

N M

L

Given: GHJN and JKLM are paralellograms. H and M are collinear. H and M are collinear.Prove: N K

CONFIDENTIAL 20

Now some problems for you to practice !

CONFIDENTIAL 21

1) BD

2) CD

3) BE

Assessment

I n llgm ABCD, AB = 17.5, DE = 18, and mBCD = 110 . Find each measure.

C D

AB

E

4) /ABC

5) /ADC

6) /DAB

CONFIDENTIAL 22

7) 8)

Find the values of x and y for which ABCD must be a parallelogram:

CONFIDENTIAL 23

9) Three vertices of llgm DFGH are D(-9,4), F(-1,5) and G(2,0). Find the coordinates of vertex H.

CONFIDENTIAL 24

10) Write a two-column proof.

Given: PSTV is a paralellogram. PQ RQ Prove: STV R

PV

TS

R

Q

CONFIDENTIAL 25

Any polygon with four sides is a quadrilateral. However, some quadrilaterals have special properties. These special

quadrilaterals have their own names.

A quadrilaterals with two pairs of parallel sides is a parallelogram. To write the name of a parallelogram, you

use the symbol □.

Properties of Parallelograms

parallelogram ABCD

□ ABCD

A B

CD

AB || CD; BC || DA

Let’s review

CONFIDENTIAL 26

Theorem Hypothesis Conclusion

If a quadrilateral is a parallelogram, then its opposite sides are congruent. (□ -> opp. sides )

Properties of Parallelograms

A B

CD

AB CDBC DA

Theorem 1:

CONFIDENTIAL 27

Proof of Theorem 1:A B

CD

1

2

3

4

Given: ABCD is a parallelogram.Proof: AB CD, BC DA

Proof:

STATEMENTS REASONS

1. ABCD is a parallelogram 1. Given2. AB CD, BC DA 2. Def. of gm3. 1 2, 3 4 3. Alt. int. s Thm.4. AC AC 4. Reflex. prop. of .5. ABC CDA 5. ASA Steps 3,46. AB CD, BC DA 6. CPCTC

CONFIDENTIAL 28

Theorem Hypothesis Conclusion

If a quadrilateral is a parallelogram, then its opposite angles are congruent. (□ -> opp. angles )

Properties of Parallelograms

A B

CD

A CB D

CONFIDENTIAL 29

Theorem Hypothesis Conclusion

If a quadrilateral is a parallelogram, then its consecutive angles are supplementary. (□ -> cons. ∕s supp.)

Properties of Parallelograms

A B

CD

mA + B 180mB + C 180mC + D 180mD + A 180

CONFIDENTIAL 30

Theorem Hypothesis Conclusion

If a quadrilateral is a parallelogram, then its diagonals bisect each other.(□ -> diags. bisect each other)

Properties of Parallelograms

A B

CD

1

2

3

4

ZAZ CZBZ DZ

CONFIDENTIAL 31

Three vertices of □ABCD are A(1,-2), B(-2,3) and D(5,-1). Find the coordinates of vertex C.

Parallelograms in the coordinate plane

B

A

D

5

60

C

-3

-2

-3

5

x

y

Since ABCD is a parallelogram, both pairs of opposite sides must

be parallel.

Step1: Graph the given points.

Step2: Find the slope of AB by counting the units from A to B.The rise from -2 to 3 is 5.The run from 1 to -2 is -3.

CONFIDENTIAL 32

B

A

D

5

60

C

-3

-2

-3

5

x

y

Step3: Start at D and count the same number of units.

The rise from -1 to 4 is 5.The run from 5 to 2 is -3.Label (2,4) as vertex C.

Step4: Use the slope formula to verify that BC || AD.

Slope of BC = 4 – 3 = 1 2 – (-2) 4

Slope of AD = -1 – (-2) = 1 5 – 1 4

The coordinates of vertex C are (2, 4).

CONFIDENTIAL 33

Using properties of Parallelograms in a proof

Given: ABCD is a parallelogram.Proof: BAD DCB, ABC CDA

Proof:

STATEMENTS REASONS

1. ABCD is a parallelogram 1. Given2. AB CD, BC DA 2. gm opp.sides 3. BD BD 3. Reflex. prop. of .4. BAF DCB 4. SSS Steps 2,35. BAD DCB 5. CPCTC6. AC AC 6. Reflex. prop. of .7. ABC CDA 7. SSS Steps 2,68. ABC CDA 8. CPCTC

B C

DA

E

A)

CONFIDENTIAL 34

You did a You did a great great job today!job today!