1. Warm-up Use the information in the diagram to solve for j.
60°
150°
3j °

2. Properties of Parallelograms 6.2
Week 1 Day 3
January 8th, 2014

3. Essential Question
How do we use properties of parallelograms?

4. Objectives: Use some properties of parallelograms.
Use properties of parallelograms in real-lie situations such as the drafting table shown in example 6.

5. In this lesson . . .
And the rest of the chapter, you will study special quadrilaterals. A parallelogram is a quadrilateral with both pairs of opposite sides parallel.
When you mark diagrams of quadrilaterals, use matching arrowheads to indicate which sides are parallel. For example, in the diagram to the right, PQ║RS and QR║SP. The symbol PQRS is read “parallelogram PQRS.”

6. Theorem 6.2 Q R
If a quadrilateral is a parallelogram, then its opposite sides are congruent.
►PQ≅RS and SP≅QR P S

7. Theorems 6.3 Q R
If a quadrilateral is a parallelogram, then its opposite angles are congruent.
P ≅ R and
Q ≅ S P S

8. Theorem 6.4 Q R
If a quadrilateral is a parallelogram, then its consecutive angles are supplementary (add up to 180°).
mP +mQ = 180°,
mQ +mR = 180°,
mR + mS = 180°,
mS + mP = 180° P S

9. Theorem 6.5 Q R
If a quadrilateral is a parallelogram, then its diagonals bisect each other.
QM ≅ SM and
PM ≅ RM P S

10. Ex. 1: Using properties of Parallelograms5 F G
FGHJ is a parallelogram. Find the unknown length. Explain your reasoning. JH JK 3 K H J b.

11. Ex. 1: Using properties of Parallelograms5 F G
FGHJ is a parallelogram. Find the unknown length. Explain your reasoning. JH JK
SOLUTION:
a. JH = FG Opposite sides of a are ≅.
JH = 5 Substitute 5 for FG. 3 K H J b.

12. Ex. 1: Using properties of Parallelograms5 F G
FGHJ is a parallelogram. Find the unknown length. Explain your reasoning. JH JK
SOLUTION:
a. JH = FG Opposite sides of a are ≅.
JH = 5 Substitute 5 for FG. 3 K H J b.
JK = GK Diagonals of a bisect each other.
JK = 3 Substitute 3 for GK

13. Ex. 2: Using properties of parallelogramsQ R
PQRS is a parallelogram.
Find the angle measure.
mR
mQ
70° P S

14. Ex. 2: Using properties of parallelogramsQ R
PQRS is a parallelogram.
Find the angle measure.
mR
mQ
a. mR = mP Opposite angles of a are ≅.
mR = 70° Substitute 70° for mP.
70° P S

15. Ex. 2: Using properties of parallelogramsQ R
PQRS is a parallelogram.
Find the angle measure.
mR
mQ
a. mR = mP Opposite angles of a are ≅.
mR = 70° Substitute 70° for mP.
mQ + mP = 180° Consecutive s of a are supplementary.
mQ + 70° = 180° Substitute 70° for mP.
mQ = 110° Subtract 70° from each side.
70° P S

16. Ex. 3: Using Algebra with ParallelogramsQ
PQRS is a parallelogram. Find the value of x.
mS + mR = 180°
3x = 180
3x = 60
x = 20
3x°
120° S R
Consecutive s of a □ are supplementary.
Substitute 3x for mS and 120 for mR.
Subtract 120 from each side.
Divide each side by 3.

17. Ex. 4: Proving Facts about Parallelograms
Given: ABCD and AEFG are parallelograms.
Prove 1 ≅ 3.
ABCD is a □. AEFG is a ▭.
1 ≅ 2, 2 ≅ 3
1 ≅ 3
Given

18. Ex. 4: Proving Facts about Parallelograms
Given: ABCD and AEFG are parallelograms.
Prove 1 ≅ 3.
ABCD is a □. AEFG is a □.
1 ≅ 2, 2 ≅ 3
1 ≅ 3
Given
Opposite s of a ▭ are ≅

19. Ex. 4: Proving Facts about Parallelograms
Given: ABCD and AEFG are parallelograms.
Prove 1 ≅ 3.
ABCD is a □. AEFG is a □.
1 ≅ 2, 2 ≅ 3
1 ≅ 3
Given
Opposite s of a ▭ are ≅
Transitive prop. of congruence.

20. Ex. 5: Proving Theorem 6.2 Given: ABCD is a parallelogram.
Prove AB ≅ CD, AD ≅ CB.
ABCD is a .
Draw BD.
AB ║CD, AD ║ CB.
ABD ≅ CDB, ADB ≅  CBD
DB ≅ DB
∆ADB ≅ ∆CBD
AB ≅ CD, AD ≅ CB
Given

21. Ex. 5: Proving Theorem 6.2 Given: ABCD is a parallelogram.
Prove AB ≅ CD, AD ≅ CB.
ABCD is a .
Draw BD.
AB ║CD, AD ║ CB.
ABD ≅ CDB, ADB ≅  CBD
DB ≅ DB
∆ADB ≅ ∆CBD
AB ≅ CD, AD ≅ CB
Given
Through any two points, there exists exactly one line.

22. Ex. 5: Proving Theorem 6.2 Given: ABCD is a parallelogram.
Prove AB ≅ CD, AD ≅ CB.
ABCD is a .
Draw BD.
AB ║CD, AD ║ CB.
ABD ≅ CDB, ADB ≅  CBD
DB ≅ DB
∆ADB ≅ ∆CBD
AB ≅ CD, AD ≅ CB
Given
Through any two points, there exists exactly one line.
Definition of a parallelogram

23. Ex. 5: Proving Theorem 6.2 Given: ABCD is a parallelogram.
Prove AB ≅ CD, AD ≅ CB.
ABCD is a .
Draw BD.
AB ║CD, AD ║ CB.
ABD ≅ CDB, ADB ≅  CBD
DB ≅ DB
∆ADB ≅ ∆CBD
AB ≅ CD, AD ≅ CB
Given
Through any two points, there exists exactly one line.
Definition of a parallelogram
Alternate Interior s Thm.

24. Ex. 5: Proving Theorem 6.2 Given: ABCD is a parallelogram.
Prove AB ≅ CD, AD ≅ CB.
ABCD is a .
Draw BD.
AB ║CD, AD ║ CB.
ABD ≅ CDB, ADB ≅  CBD
DB ≅ DB
∆ADB ≅ ∆CBD
AB ≅ CD, AD ≅ CB
Given
Through any two points, there exists exactly one line.
Definition of a parallelogram
Alternate Interior s Thm.
Reflexive property of congruence

25. Ex. 5: Proving Theorem 6.2 Given: ABCD is a parallelogram.
Prove AB ≅ CD, AD ≅ CB.
ABCD is a .
Draw BD.
AB ║CD, AD ║ CB.
ABD ≅ CDB, ADB ≅  CBD
DB ≅ DB
∆ADB ≅ ∆CBD
AB ≅ CD, AD ≅ CB
Given
Through any two points, there exists exactly one line.
Definition of a parallelogram
Alternate Interior s Thm.
Reflexive property of congruence
ASA Congruence Postulate

26. Ex. 5: Proving Theorem 6.2 Given: ABCD is a parallelogram.
Prove AB ≅ CD, AD ≅ CB.
ABCD is a .
Draw BD.
AB ║CD, AD ║ CB.
ABD ≅ CDB, ADB ≅  CBD
DB ≅ DB
∆ADB ≅ ∆CBD
AB ≅ CD, AD ≅ CB
Given
Through any two points, there exists exactly one line.
Definition of a parallelogram
Alternate Interior s Thm.
Reflexive property of congruence
ASA Congruence Postulate
CPCTC

27. Ex. 6: Using parallelograms in real life
FURNITURE DESIGN. A drafting table is made so that the legs can be joined in different ways to change the slope of the drawing surface. In the arrangement below, the legs AC and BD do not bisect each other. Is ABCD a parallelogram?

28. Ex. 6: Using parallelograms in real life
FURNITURE DESIGN. A drafting table is made so that the legs can be joined in different ways to change the slope of the drawing surface. In the arrangement below, the legs AC and BD do not bisect each other. Is ABCD a parallelogram?
ANSWER: NO. If ABCD were a parallelogram, then by Theorem 6.5, AC would bisect BD and BD would bisect AC. They do not, so it cannot be a parallelogram.

29. Assignment:
pp. 314 #22-33 (even)