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 $\mathrm{△}ABC\cong \mathrm{△}ADC$ 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, $\mathrm{△}ABC\cong \mathrm{△}ADC$ 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, $\mathrm{△}ABE\cong \mathrm{△}CDE$.

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 $\mathrm{△}AED\cong \mathrm{△}CEB$ and $\mathrm{△}CED\cong \mathrm{△}AEB$.

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