Question

6-4: proving a quadrilateral is a parallelogram
which condition is sufficient to show that quadrilateral defg is a parallelogram?
a ∠gde is supplementary to ∠def
b. ∠gde is supplementary to ∠efg
c. ∠gde is supplementary to ∠def and ∠fgd
d. ∠gde is congruent to ∠fgd

Explanation:

To determine which condition shows DEFG is a parallelogram, we use properties of parallelograms and supplementary angles. A parallelogram has opposite sides parallel, and consecutive angles between parallel sides are supplementary.

• Option A: If \( \angle GDE \) is supplementary to \( \angle DEF \), these are consecutive angles (along side \( DE \)). Supplementary consecutive angles imply \( DG \parallel EF \) (since same - side interior angles are supplementary, lines are parallel). But we need another pair of parallel sides. However, let's check other options.
• Option B: \( \angle GDE \) and \( \angle EFG \) are not consecutive or related in a way that directly implies parallel sides. \( \angle GDE \) is at vertex \( D \) and \( \angle EFG \) is at vertex \( F \), so this doesn't help establish parallel sides.
• Option C: If \( \angle GDE \) is supplementary to \( \angle DEF \), then \( DG \parallel EF \) (same - side interior angles supplementary). If \( \angle GDE \) is supplementary to \( \angle FGD \), then \( DE \parallel FG \) (same - side interior angles supplementary). If both pairs of opposite sides are parallel (\( DG \parallel EF \) and \( DE \parallel FG \)), then DEFG is a parallelogram.
• Option D: If \( \angle GDE\cong\angle FGD \), this implies \( DE \parallel FG \) (alternate interior angles congruent), but we don't know about the other pair of sides. So this is not sufficient.

Answer:

C. \( \angle GDE \) is supplementary to \( \angle DEF \) and \( \angle FGD \)