# 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 $\mathrm{\u25b3}ABC\cong \mathrm{\u25b3}ADC$ by SSS, since by assumption^{} $AB=CD$ and $AD=BC$, and the two triangles^{} share a third side.

By CPCTC, it follows that $\mathrm{\angle}BAC\cong \mathrm{\angle}DCA$ and that $\mathrm{\angle}BCA\cong \mathrm{\angle}DAC$. 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: $\mathrm{\angle}BAC\cong \mathrm{\angle}DCA$. Then by SAS, $\mathrm{\u25b3}ABC\cong \mathrm{\u25b3}ADC$ 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: $\mathrm{\angle}ABD\cong \mathrm{\angle}BDC$ and $\mathrm{\angle}BAC\cong \mathrm{\angle}ACD$. Then by ASA, $\mathrm{\u25b3}ABE\cong \mathrm{\u25b3}CDE$.

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 $\mathrm{\angle}AED\cong \mathrm{\angle}AEB$ and $\mathrm{\angle}CED\cong \mathrm{\angle}AEB$. Thus, by SAS we have that $\mathrm{\u25b3}AED\cong \mathrm{\u25b3}CEB$ and $\mathrm{\u25b3}CED\cong \mathrm{\u25b3}AEB$.

By CPCTC it follows that $AB=CD$ and that $AD=BC$. By Theorem 1, $ABCD$ is a parallelogram.

∎

Title | proof of parallelogram theorems |
---|---|

Canonical name | ProofOfParallelogramTheorems |

Date of creation | 2013-03-22 17:15:48 |

Last modified on | 2013-03-22 17:15:48 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 9 |

Author | rm50 (10146) |

Entry type | Proof |

Classification | msc 51M04 |

Classification | msc 51-01 |

Related topic | Parallelogram |