# proof of parallelogram theorems

###### Theorem 1.

The opposite sides of a parallelogram are congruent.

###### Proof.

This was proved in the parent (http://planetmath.org/ParallelogramTheorems) 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 $\overline{AC}$.

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 $\overline{AB}$ and $\overline{CD}$ are parallel, and that $\overline{AD}$ and $\overline{BC}$ 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 $\overline{AB}$ and $\overline{CD}$ are parallel, and draw $\overline{AC}$.

Since $\overline{AB}$ and $\overline{CD}$ 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 2, 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 $\overline{AC}$ and $\overline{BD}$, intersecting at $E$.

Since $ABCD$ is a parallelogram, we have that $AB=CD$. In addition, $\overline{AB}$ and $\overline{CD}$ 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 1, $ABCD$ is a parallelogram.

Title proof of parallelogram theorems ProofOfParallelogramTheorems 2013-03-22 17:15:48 2013-03-22 17:15:48 rm50 (10146) rm50 (10146) 9 rm50 (10146) Proof msc 51M04 msc 51-01 Parallelogram