Proving that a quadrilateral is a parallelogram

Geometry, grade 9 Lesson 13 of 21Learning

Proving that a quadrilateral is a parallelogram

The parallelogram’s properties

In the previous lesson you learned the parallelogram’s properties: its opposite sides are equal, its opposite angles are equal, and its diagonals bisect each other.

Today the other way round: a quadrilateral with one of these properties is a parallelogram.

The converse theorems

The definition is a way too: a quadrilateral whose opposite sides are parallel is a parallelogram.

Theorem: a quadrilateral is a parallelogram if any one of these holds in it:

both pairs of opposite sides are equal;
two opposite sides are parallel and equal;
the diagonals bisect each other;
both pairs of opposite angles are equal.

Proof: two pairs of equal opposite sides

Given: $\displaystyle {AB}={CD}$ and $\displaystyle {BC}={DA}$ .
Prove: $\displaystyle ABCD$ is a parallelogram.

The proof below draws the diagonal $\displaystyle AC$ and shows that triangles $\displaystyle ABC$ and $\displaystyle CDA$ are congruent. Their equal angles then show that the opposite sides are parallel.

	Claim	Reason
1	$\displaystyle {AB}={CD}$	Given
2	$\displaystyle {BC}={DA}$	Given
3	$\displaystyle {AC}={CA}$	A common side
4	$\displaystyle \triangle ABC\cong\triangle CDA$	(1), (2), (3); the congruence theorem SSS
5	$\displaystyle {\angle BAC}={\angle DCA}$	(4); in congruent triangles, equal sides face equal angles
6	$\displaystyle AB\parallel DC$	(5); if alternate angles are equal, the lines are parallel
7	$\displaystyle {\angle BCA}={\angle DAC}$	(4); in congruent triangles, equal sides face equal angles
8	$\displaystyle AD\parallel BC$	(7); if alternate angles are equal, the lines are parallel
9	$\displaystyle ABCD$ is a parallelogram	(6), (8); the definition – both pairs of opposite sides are parallel

Proof: a pair of sides parallel and equal

Careful: the equal sides must be the same pair as the parallel ones. In an isosceles trapezoid one pair of sides is parallel and the other pair is equal, and it is not a parallelogram.

Given: $\displaystyle AB\parallel DC$ and $\displaystyle {AB}={DC}$ .
Prove: $\displaystyle ABCD$ is a parallelogram.

The proof below draws the diagonal $\displaystyle AC$ too; this time it shows the triangles congruent by SAS.

	Claim	Reason
1	$\displaystyle AB\parallel DC$	Given
2	$\displaystyle {\angle BAC}={\angle DCA}$	(1); alternate angles between parallel lines are equal
3	$\displaystyle {AB}={CD}$	Given
4	$\displaystyle {AC}={CA}$	A common side
5	$\displaystyle \triangle ABC\cong\triangle CDA$	(2), (3), (4); the congruence theorem SAS
6	$\displaystyle {\angle BCA}={\angle DAC}$	(5); in congruent triangles, equal sides face equal angles
7	$\displaystyle AD\parallel BC$	(6); if alternate angles are equal, the lines are parallel
8	$\displaystyle ABCD$ is a parallelogram	(1), (7); the definition – both pairs of opposite sides are parallel

Proof: equal opposite angles

In quadrilateral $\displaystyle ABCD$ , $\displaystyle {\angle A}={\angle C}$ and $\displaystyle {\angle B}={\angle D}$ .
The proof below shows that each pair of adjacent angles adds up to $\displaystyle 180^{\circ}$ , so the opposite sides are parallel:

The angles of a quadrilateral add up to $\displaystyle 360^{\circ}$ , so $\displaystyle {2\angle A+2\angle B}={360^{\circ}}$ , that is $\displaystyle {\angle A+\angle B}={180^{\circ}}$ .
$\displaystyle \angle A$ and $\displaystyle \angle B$ are co-interior angles between the lines $\displaystyle AD$ and $\displaystyle BC$ . They add up to $\displaystyle 180^{\circ}$ , so $\displaystyle AD\parallel BC$ .
In the same way $\displaystyle {\angle A+\angle D}={180^{\circ}}$ , so $\displaystyle AB\parallel DC$ .

A parallelogram in a coordinate system

The points $\displaystyle A$ , $\displaystyle B$ , $\displaystyle C$ and $\displaystyle D$ are the vertices of quadrilateral $\displaystyle ABCD$ .
It is a parallelogram:

$\displaystyle AB$ and $\displaystyle DC$ are horizontal, so $\displaystyle AB\parallel DC$ ;
$\displaystyle {AB}={5-1}={4}$ and $\displaystyle {DC}={7-3}={4}$ , so $\displaystyle {AB}={DC}$ ;
a pair of opposite sides is parallel and equal, so $\displaystyle ABCD$ is a parallelogram.

Exercise 1

Is quadrilateral $\displaystyle ABCD$ sure to be a parallelogram? Give the reason.

$\displaystyle {AB}={CD}={7}$ and $\displaystyle {BC}={AD}={4}$ .

$\displaystyle AB\parallel DC$ and $\displaystyle {AD}={BC}$ .

$\displaystyle {\angle A}={\angle C}={115^{\circ}}$ and $\displaystyle {\angle B}={65^{\circ}}$ .

The diagonals cross at the point $\displaystyle O$ .
$\displaystyle {AO}={OC}={5}$ , $\displaystyle {BO}={3}$ and $\displaystyle {OD}={4}$ .

Exercise 2

The diagonals of quadrilateral $\displaystyle ABCD$ cross at the point $\displaystyle O$ .
$\displaystyle {AO}={OC}$ and $\displaystyle {BO}={OD}$ .
Prove: $\displaystyle ABCD$ is a parallelogram.

Exercise 3

$\displaystyle ABCD$ is a parallelogram.
The points $\displaystyle E$ and $\displaystyle F$ are on the diagonal $\displaystyle BD$ , so that $\displaystyle {BE}={DF}$ .
Prove: $\displaystyle AECF$ is a parallelogram.

Exercise 4

$\displaystyle ABCD$ is a parallelogram.
The side $\displaystyle AB$ is extended beyond $\displaystyle B$ to the point $\displaystyle E$ , so that $\displaystyle {BE}={AB}$ .
Prove: $\displaystyle BECD$ is a parallelogram.

Formulas: geometry, grade 9

Open as a page

Angles and sides in a triangle

The sum of the angles of a triangle
$\displaystyle \alpha+\beta+\gamma=180^{\circ}$
An exterior angle $\displaystyle \delta$ equals the sum of the two interior angles not next to it
$\displaystyle \delta$ $\displaystyle =$ $\displaystyle \alpha+\beta$
The larger side faces the larger angle, and the other way round
$\displaystyle a>b\ \Leftrightarrow\ \alpha>\beta$

The right triangle

Pythagoras’ theorem – $\displaystyle a$ , $\displaystyle b$ the legs, $\displaystyle c$ the hypotenuse
$\displaystyle a^{2}+b^{2}$ $\displaystyle =$ $\displaystyle c^{2}$
The leg $\displaystyle a$ facing a $\displaystyle 30^{\circ}$ angle is half the hypotenuse $\displaystyle c$
$\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{2}c$
The median $\displaystyle m_c$ to the hypotenuse is half the hypotenuse $\displaystyle c$
$\displaystyle m_c$ $\displaystyle =$ $\displaystyle \frac{1}{2}c$

Angles between parallel lines

Corresponding angles are equal, and so are alternate angles
$\displaystyle \alpha=\beta$
Co-interior angles
$\displaystyle \alpha+\beta=180^{\circ}$

Angles in a quadrilateral

The sum of the angles of a quadrilateral
$\displaystyle \alpha+\beta+\gamma+\delta=360^{\circ}$
In a trapezoid: the two angles beside a leg
$\displaystyle \alpha+\beta=180^{\circ}$

Areas

The area of a triangle – $\displaystyle h_a$ the height to the side $\displaystyle a$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{a\cdot h_a}{2}$
The area of a parallelogram – $\displaystyle h_a$ the height to the side $\displaystyle a$
$\displaystyle S$ $\displaystyle =$ $\displaystyle a\cdot h_a$
The area of a rectangle
$\displaystyle S$ $\displaystyle =$ $\displaystyle a\cdot b$
The area of a square
$\displaystyle S$ $\displaystyle =$ $\displaystyle a^{2}$
The area of a trapezoid – $\displaystyle a$ , $\displaystyle b$ the bases, $\displaystyle h$ the height
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{(a+b)\cdot h}{2}$
The area of a rhombus or a kite – $\displaystyle d_1$ , $\displaystyle d_2$ the diagonals
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{d_1\cdot d_2}{2}$
The ratio of the areas of similar triangles – $\displaystyle k$ the ratio of similarity
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$