Theorem 1   The opposite sides of a parallelogram are congruent.

Proof. This was proved in the parent article.

Theorem 2   If both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram.

Proof. Suppose $ABCD$ is the given parallelogram, and draw $\ol{AC}$ Then $\triangle ABC\cong\triangle ADC$ by SSS, since by assumption $AB=CD$ and $AD=BC$ and the two triangles share a third side.

By CPCTC, it follows that $\angle BAC\cong\angle DCA$ and that $\angle BCA\cong \angle DAC$ But the theorems about corresponding angles in transversal cutting then imply that $\ol{AB}$ and $\ol{CD}$ are parallel, and that $\ol{AD}$ and $\ol{BC}$ are parallel. Thus $ABCD$ is a parallelogram.

Theorem 3   If one pair of opposite sides of a quadrilateral are both parallel and congruent, the quadrilateral is a parallelogram.

Proof. Again let $ABCD$ be the given parallelogram. Assume $AB=CD$ and that $\ol{AB}$ and $\ol{CD}$ are parallel, and draw $\ol{AC}$ Since $\ol{AB}$ and $\ol{CD}$ are parallel, it follows that the alternate interior angles are equal: $\angle BAC\cong \angle DCA$ Then by SAS, $\triangle ABC\cong \triangle ADC$ since they share a side.

Again by CPCTC we have that $BC=AD$ so both pairs of sides of the quadrilateral are congruent, so by Theorem , the quadrilateral is a parallelogram.

Theorem 4   The diagonals of a parallelogram bisect each other.

Proof. Let $ABCD$ be the given parallelogram, and draw the diagonals $\ol{AC}$ and $\ol{BD}$ intersecting at $E$ Since $ABCD$ is a parallelogram, we have that $AB=CD$ In addition, $\ol{AB}$ and $\ol{CD}$ are parallel, so the alternate interior angles are equal: $\angle ABD\cong \angle BDC$ and $\angle BAC\cong \angle ACD$ Then by ASA, $\triangle ABE\cong \triangle CDE$

By CPCTC we see that $AE=CE$ and $BE=DE$ proving the theorem.

Theorem 5   If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.

Proof. Let $ABCD$ be the given quadrilateral, and let its diagonals intersect in $E$ Then by assumption, $AE=EC$ and $DE=EB$ But also vertical angles are equal, so $\angle AED\cong \angle AEB$ and $\angle CED\cong \angle AEB$ Thus, by SAS we have that $\triangle AED\cong \triangle CEB$ and $\triangle CED\cong \triangle AEB$

By CPCTC it follows that $AB=CD$ and that $AD=BC$ By Theorem , $ABCD$ is a parallelogram.