Properties of Parallelograms
State and prove the six properties of a parallelogram (opposite sides parallel and congruent, opposite angles congruent, consecutive angles supplementary, diagonals bisect each other). Apply in proof and calculation.
A lesson of Geometry, in Quadrilaterals and Polygons, about 10 minutes.
By the end of this lesson, you will be able to state and apply all five properties of parallelograms, prove theorems about parallelograms using congruent triangles, and determine if a quadrilateral is a parallelogram.
The Big Idea
A parallelogram is a quadrilateral with two pairs of parallel sides. From this one definition, five powerful properties follow — and all five can be proven using the triangle congruence tools from earlier units. This lesson is a demonstration of how a single geometric definition generates a rich structure.
Definition
$ABCD$ is a parallelogram if $\overline{AB} \parallel \overline{CD}$ and $\overline{BC} \parallel \overline{AD}$.
Five Properties of Parallelograms
Property 1: Opposite sides are congruent.
$AB = CD$ and $BC = AD$
Property 2: Opposite angles are congruent.
$\angle A \cong \angle C$ and $\angle B \cong \angle D$
Property 3: Consecutive angles are supplementary.
$\angle A + \angle B = 180°$ (and same for all consecutive pairs)
Property 4: Diagonals bisect each other.
The diagonals intersect at their mutual midpoint.
Property 5: Each diagonal divides the parallelogram into two congruent triangles.
$\triangle ABC \cong \triangle CDA$ (by SSS or AAS)
Proof of Property 4
Given: $ABCD$ is a parallelogram with diagonals $\overline{AC}$ and $\overline{BD}$ intersecting at $E$.
Prove: $AE = CE$ and $BE = DE$ (diagonals bisect each other).
$\overline{AB} \parallel \overline{CD}$ → $\angle ABE \cong \angle CDE$ (alternate interior angles, transversal $\overline{BD}$)
$\overline{AB} \parallel \overline{CD}$ → $\angle BAE \cong \angle DCE$ (alternate interior angles, transversal $\overline{AC}$)
$AB = CD$ (Property 1)
$\triangle ABE \cong \triangle CDE$ by ASA
Therefore $AE = CE$ and $BE = DE$ by CPCTC. ∎
Example 1 — Finding Measures
Parallelogram $ABCD$: $AB = 12$, $BC = 8$, $\angle A = 65°$.
By Property 1: $CD = 12$, $AD = 8$
By Property 2: $\angle C = 65°$
By Property 3: $\angle B = 180° - 65° = 115°$, $\angle D = 115°$
Example 2 — Diagonals
In parallelogram $ABCD$, diagonals intersect at $E$. $AE = 3x - 1$ and $CE = x + 7$. Find $AC$.
Diagonals bisect: $AE = CE$
$$3x - 1 = x + 7 \implies 2x = 8 \implies x = 4$$