1. Conditions for Parallelograms Students will be able to show that a given quadrilateral is a parallelogram

2. Unit G2 Characteristics or Conditions In the last section we learned five characteristics of a parallelogram. We use these when we already know the figure is a parallelogram. In this section, we will learn what conditions will prove that a quadrilateral is a parallelogram. If we have one of these, then we know we have a parallelogram.

3. Unit G3 Condition 1: The Definition of a Parallelogram If both pairs of opposite sides are parallel, then the quadrilateral is a parallelogram. A D C B

4. Unit G4 Condition 2: One Pair of Opposite Sides If one pair of opposite sides is both parallel and congruent, then the quadrilateral is a parallelogram. A D C B

5. Unit G5 Condition 3: Both Pairs of Opposite Sides If both pairs of opposite sides are congruent, then the quadrilateral is a parallelogram. A D C B

6. Unit G6 Condition 4: Both Pairs of Opposite Angles If both pairs of opposite angles are congruent, then the quadrilateral is a parallelogram. A B C D

7. Unit G7 Condition 5: Consecutive Angles If an angle is supplementary to both of its consecutive angles, then the quadrilateral is a parallelogram. A B C D xº (180 – x)º

8. Unit G8 Condition 6: Bisecting Diagonals If the diagonals bisect each other, then the quadrilateral is a parallelogram. A D C B

9. Unit G9 Example 1: Determine if each quadrilateral is a parallelogram Yes. The 73° angle is supplementary to both its consecutive angles. No. Only one pair of opposite angles are congruent. The conditions for a parallelogram are not met.

10. Unit G10 Example 2: Determine if each quadrilateral is a parallelogram Yes. Since angles 1 and 2 are congruent by the 3 rd Angle Theorem, both pairs of opposite angles of the quadrilateral are congruent. No. Two pairs of consecutive sides are congruent. None of the sets of conditions for a parallel- ogram are met. 1 2

11. Unit G11 Example 3: Find the values of a and b that would make the quadrilateral a parallelogram. 15a – 11 = 10a + 4 5a = 15 a = 3 5b + 6 = 8b – 21 27 = 3b 9 = b