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.”

If a quadrilateral is a parallelogram, then its opposite sides are congruent.

►PQ RS and ≅SP QR≅

P

Q R

S

If a quadrilateral is a parallelogram, then its opposite angles are congruent.

P ≅ R andQ ≅ S

P

Q R

S

6.4—If a quadrilateral is a parallelogram, then its consecutive angles are supplementary (add up to 180°).

mP +mQ = 180°,mQ +mR = 180°, mR + mS = 180°, mS + mP = 180°

P

Q R

S

6.5—If a quadrilateral is a parallelogram, then its diagonals bisect each other.

QM ≅ SM and PM ≅ RM

P

Q R

S

FGHJ is a parallelogram. Find the unknown length. Explain your reasoning.a. JHb. JK

F G

J H

K

5

3

FGHJ is a parallelogram. Find the unknown length. Explain your reasoning.a. JHb. JK

SOLUTION:a. JH = FG Opposite

sides of a are ≅. b. JH = 5 Substitute 5

for FG.

F G

J H

K

5

3

PQRS is a parallelogram. Find the angle measure.a. mRb. mQ

P

RQ

70°

S

PQRS is a parallelogram. Find the angle measure.a. mRb. mQ

Opposite angles of a are ≅.Substitute 70° for mP.

mR = mP mR = 70°

P

RQ

70°

S

PQRS is a parallelogram. Find the angle measure.a. mRb. mQa. mR = mP Opposite angles of a are ≅. mR = 70° Substitute 70° for mP.

b. mQ + mP = 180° Consecutive s of a are supplementary.

mQ + 70° = 180° Substitute 70° for mP. mQ = 110° Subtract 70° from each

side.

P

RQ

70°

S

PQRS is a parallelogram. Find the value of x.

mS + mR = 180°3x + 120 = 180

3x = 60x = 20

Consecutive s of a □ are supplementary.

Substitute 3x for mS and 120 for mR.

Subtract 120 from each side.Divide each side by 3.

S

QP

R3x° 120°

FGHJ is a parallelogram. Find the unknown length. Explain your reasoning.a. JHb. JK

SOLUTION:a. JH = FG Opposite

sides of a are ≅. JH = 5 Substitute 5

for FG.

F G

J H

K

5

3

b. JK = GK Diagonals of a bisect each other.

JK = 3