Proving a Quadrilateral Is a Parallelogram: Practice

The three complete side or angle conditions are sufficient tests. One midpoint condition leaves the other diagonal unrestricted. For example, the perpendicular-diagonal construction with halves $6,6$ and $5,3$ satisfies one bisection and fails the other.

Use the quadrilateral angle sum:

Thus the second opposite pair is congruent too, completing the angle test.

The proposed counterexample has equal angles $A,C$ but unequal angles $B,D$, which excludes a parallelogram. It exists: two right triangles on opposite sides of a common hypotenuse can have their other angles $30^\circ,60^\circ$, joined so the combined angles at the hypotenuse endpoints are $60^\circ$ and $120^\circ$.

All three other choices have both opposite-angle pairs equal, so they are parallelograms.

A parallelogram must have congruent opposite sides. Unequal values $12$ and $6$ for $AB,CD$ contradict that necessary property. Congruent sloping sides, an obtuse angle, and unequal diagonals can all occur in parallelograms.

The first midpoint equality gives

The second gives

At this value, both half-diagonal pairs are positive and equal within their own diagonal.

To prove $AD\parallel BC$, use $BD$ as transversal. Its alternate interior angles are $\angle ADB$ and $\angle CBD$, corresponding under the given triangle order.

The equality at $B,D$ involving $AB,CD$ concerns the parallel pair already given.

Triangles $BEC,DEA$ have equal half-diagonals around equal vertical angles, so SAS applies. Their corresponding sides give $BC=DA$. Together with the first opposite-side equality, this completes the side test.

Points $A,E,C$ are collinear, so $AEC$ is not a triangle.

The first two facts in the incompatible collection prove a parallelogram by the one-pair test. Its diagonals must then bisect each other, contradicting $AE\ne EC$.

Each other collection describes valid complete parallelogram evidence with positive lengths or valid interior angles.

The two facts attach parallelism and congruence to the same opposite-side pair, so they suffice. Either one alone can hold in a quadrilateral that is not a parallelogram. The three-fact option is sufficient but contains an extra fact.

For $AC$ to be bisected,

which requires $v=6$. For $BD$ to be bisected,

which requires $v=5$. The same parameter cannot take both values, so this test has no solution.

The first sum concerns the same-side interior angles along transversal $AB$, so it proves $AD\parallel BC$.

The second concerns the same-side interior angles along transversal $BC$, so it proves $AB\parallel CD$.

Both parallel pairs establish the definition; the angles need not all be right.

Adding $AB\parallel CD$ makes the same opposite-side pair both parallel and congruent, a sufficient test.

Perpendicular opposite sides do not meet this test. Perpendicular diagonals or a further adjacent-side equality do not ensure the second opposite-side equality or a parallel pair.