Properties of a Parallelogram

A parallelogram is the most important named quadrilateral. The single defining property , both pairs of opposite sides parallel , forces a cascade of beautiful consequences: equal opposite sides, equal opposite angles, and diagonals that bisect each other. This lesson proves the cascade.

Definitions

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

In parallelogram $ABCD$:

• Sides $AB$ and $CD$ are opposite (and parallel).
• Sides $AD$ and $BC$ are opposite (and parallel).
• The diagonals are $AC$ and $BD$.

Properties (and proofs)

Property 1. Opposite sides of a parallelogram are equal.

Proof. In parallelogram $ABCD$, draw diagonal $AC$. Then $AB \parallel DC$ (definition), so $\angle BAC = \angle DCA$ (alternate-interior angles). Also $AD \parallel BC$ (definition), so $\angle DAC = \angle BCA$ (alternate-interior angles). And $AC = CA$ (common). By ASA, $\triangle ABC \cong \triangle CDA$. Hence $AB = DC$ and $BC = DA$. Q.E.D.

Property 2. Opposite angles of a parallelogram are equal.

Proof. In the above congruence, by CPCT, $\angle B = \angle D$. Similarly by drawing the other diagonal $BD$, we get $\angle A = \angle C$. Q.E.D.

Property 3. Consecutive angles of a parallelogram are supplementary.

Proof. Since $AB \parallel DC$ with transversal $AD$, co-interior angles $\angle A$ and $\angle D$ sum to $180^\circ$. Similarly all consecutive pairs. Q.E.D.

Property 4. The diagonals of a parallelogram bisect each other.

Proof. In parallelogram $ABCD$, let the diagonals $AC$ and $BD$ intersect at $O$. Consider $\triangle AOB$ and $\triangle COD$. We have $AB = CD$ (Property 1), $\angle OAB = \angle OCD$ (alternate-interior angles, since $AB \parallel CD$ with transversal $AC$), and $\angle ABO = \angle CDO$ (alternate-interior angles, since $AB \parallel CD$ with transversal $BD$). By ASA, $\triangle AOB \cong \triangle COD$. Hence $OA = OC$ and $OB = OD$ by CPCT. Q.E.D.

So the four key properties of every parallelogram are: opposite sides equal, opposite angles equal, consecutive angles supplementary, diagonals bisecting each other.

A useful consequence

Each diagonal divides a parallelogram into two congruent triangles. This is exactly Property 1's proof. So problems involving parallelograms often reduce to congruence-of-triangle proofs by drawing a diagonal.

Three classical applications

Application 1. In parallelogram $ABCD$, $\angle A = 70^\circ$. Find the other three angles. By Property 2, $\angle C = 70^\circ$. By Property 3, $\angle B = 180 - 70 = 110^\circ$, and $\angle D = 110^\circ$.

Application 2. In parallelogram $ABCD$ with $AB = 5, AD = 7$, find the perimeter. Opposite sides equal, so $CD = 5$ and $BC = 7$. Perimeter $= 2 \cdot 5 + 2 \cdot 7 = 24$.

Application 3. The diagonals of a parallelogram are $10$ and $14$. Find $OA, OB, OC, OD$. Diagonals bisect each other (Property 4): $OA = OC = 5$, $OB = OD = 7$.

Worked examples

Example 1. In parallelogram $PQRS$, $\angle P = 65^\circ$. Find $\angle Q, \angle R, \angle S$.

$\angle R = 65^\circ$ (opposite, equal). $\angle Q = 180 - 65 = 115^\circ$ (consecutive, supplementary). $\angle S = 115^\circ$.

Example 2. $ABCD$ is a parallelogram with $AB = 6$ and perimeter $20$. Find $AD$.

$2(AB + AD) = 20 \Rightarrow AB + AD = 10 \Rightarrow AD = 4$.

Example 3. In parallelogram $ABCD$, the diagonals meet at $O$. $OA = 4, OB = 5$. Find $OC, OD, AC, BD$.

Diagonals bisect: $OC = OA = 4$, $OD = OB = 5$. $AC = 8, BD = 10$.

Example 4. In parallelogram $ABCD$, $\angle B$ is $20^\circ$ more than $\angle A$. Find $\angle A$.

$\angle A + \angle B = 180^\circ$ and $\angle B = \angle A + 20$. So $\angle A + \angle A + 20 = 180 \Rightarrow 2 \angle A = 160 \Rightarrow \angle A = 80^\circ$.

Example 5. In parallelogram $ABCD$, prove $\angle A + \angle C = \angle B + \angle D = 180^\circ$.

By Property 2, $\angle A = \angle C$. Also $\angle A + \angle B = 180^\circ$ (Property 3). Adding $\angle A = \angle C$ to itself: $2\angle A = 360 - 2\angle B$, so $\angle A + \angle B = 180^\circ$ (same as before). Hence $\angle A + \angle C = 2\angle A$. To get this $= 180^\circ$, we need $\angle A = 90^\circ$, which is the rectangle case. So in general, $\angle A + \angle C$ is just $2\angle A$, not $180^\circ$. Correction: The intended claim is that consecutive angles sum to $180^\circ$ (already Property 3). The opposite angles are equal, not supplementary. So $\angle A + \angle B = 180^\circ$ holds always; $\angle A + \angle C = 180^\circ$ holds only in rectangles. Be careful.

Try it yourself

1. State the four properties of a parallelogram.
2. In parallelogram $ABCD$, $\angle A = 110^\circ$. Find $\angle B, \angle C, \angle D$.
3. In parallelogram $PQRS$ with $PQ = 7, QR = 5$, find the perimeter.
4. Prove that the diagonals of a parallelogram bisect each other.
5. In parallelogram $ABCD$, the diagonals intersect at $O$, with $OA = 3$ and $OB = 4$. Find $AC$ and $BD$.
6. In parallelogram $ABCD$, $\angle A - \angle B = 30^\circ$. Find both.
7. Prove that consecutive angles of a parallelogram are supplementary.
8. Show that a parallelogram with one right angle is a rectangle.
9. Two adjacent sides of a parallelogram have lengths $5$ and $9$. Find its perimeter.
10. In parallelogram $ABCD$, $AB = 2x + 1, CD = 3x - 4$. Find $x$.

Pitfalls / Insight

• Opposite vs. consecutive. Opposite angles are equal; consecutive angles are supplementary.
• Property 4 says diagonals bisect each other , they need not be equal (rectangles are special).
• Always identify which pair of sides is parallel. The labelling $ABCD$ goes around the figure.

Insight. Once you can spot a parallelogram, you get four equalities for free , that is the entire point. These four properties make parallelograms the most useful named quadrilateral, and they form the basis for the special cases (rectangles, rhombi, squares) of the next lesson.