Parallelograms & Related Quadrilaterals

General Properties of a Parallelogram:

• The opposite sides are parallel
• The opposite sides are equal
• The opposite angles are equal
• The diagonals bisectBisect means dividing into two equal parts. It means to divide a geometric figure such as a line, an angle or any other shape into two congruent parts (or two parts... More one another
• The formula for finding the sum of the measure of the interior angles is (n – 2) × 180º
• There are 4 sides in a parallelogram; therefore, the sum of the interior angles of a parallelogram is:
  • (4 – 2) × 180º
  • = 360º

Types of Parallelogram:

1. Rhomboid:

This is a quadrilateral which has both pairs of opposite sides parallel.

A rhomboid is often referred to as a parallelogram.

Properties of a Rhomboid:

• The opposite sides are parallel
• The opposite sides are equal |AD| = |BC|, |AB| = |DC|
• The opposite angles are equal ∠A = ∠C, ∠D = ∠B
  • y₂ = y₁, x₂ = x₁ (alternate angles)
  • ΔACB ≅ ΔACD (congruent triangles)
• The diagonals bisect one another
• Consecutive angles are supplementarySupplementary angles are two angles whose sum is 180 degrees. More (meaning they add up to 180º)

2. Rhombus:

Properties of a Rhombus:

• All four sides are equal
• The opposite sides are parallel
• The opposite angles are equal (∠S = ∠Q, ∠P = ∠R)
• The diagonals bisect the angles
• The diagonals intersect at 90°, meaning they are perpendicular to each other.
• Adjacent angles are supplementary (e.g., ∠S + ∠R = 180°).

3. Rectangle:

Properties of a Rectangle:

• All of the properties of a parallelogram are found
• All four angles are right angles (90º)
• The diagonals are equal

Kite:

Properties of a Kite:

• Diagonals bisect at right angles
• The longer diagonal bisects the shorter diagonal
• Only one pair of opposite angles is equal, i.e. ∠ABC = ∠ADC
• Each pair of adjacent sides is equal

Square:

Properties of a Square:

• All of the properties of a rhombus are found
• All four angles are right angles
• The diagonals are equal
• The diagonals bisect each other at right angles

Theorems Related to Angles of a Parallelogram:

Given: Parallelogram ADCB

To prove:

1. \( \scriptsize \overline{AD} = \overline{BC}, \: and \: \overline{DC} = \overline{AB}. \)
2. Opposite angles of a parallelogram are equal.
3. Consecutive angles of a parallelogram are supplementary.

Proof:

Draw a diagonal DB.

In the parallelogram ADCB, diagonal DB divides the parallelogram into two triangles, ΔDAB and ΔDCB.

Thus, the diagonal DB acts as a transversal and is common to both triangles.

On comparing triangles ΔDAB and ΔDCB, we have:

• x₁ = x₂ (alternate angles, \( \scriptsize \overline{AB} || \overline{DC} \))
• y₁ = y₂ (alternate angles, \( \scriptsize \overline{AD} || \overline{BC} \))

Therefore, ΔDAB and ΔDCB are congruent. Two angles and a side are equal to two angles and the corresponding side of the other (i.e. ASA or AAS).

(1)

Hence,

⇒ \( \scriptsize \overline{AD} = \overline{BC}\)

⇒ \( \scriptsize \overline{DC} = \overline{AB} \)

(2)

∠D = ∠B

Because:

∠D = x₁ + y₁ and ∠B = x₂ + y₂

but x₁ + y₁ = x₂ + y₂

∴ ∠D = ∠B

Similarly, we can show that ∠A = ∠C

∴ Opposite angles of a parallelogram are equal.

(3)

We know that the interior angles in a quadrilateral add up to 360º

∴ ∠D + ∠B + ∠A + ∠C = 360º

Now, we can substitute ∠D with ∠B and ∠A with ∠C

∴ 2(∠D) + 2(∠C) = 360º

⇒ 2(∠D + ∠C) = 360º

⇒ ∠D + ∠C = \( \frac{360º}{2} \\ = \scriptsize 180º\)

 This shows that the consecutive angles are supplementary. 

 It means that \( \scriptsize \overline{AD} || \overline{BC} \)

Example 3.2.1:

Example 3.2.2:

In the diagram below, find i and j.

Solution

Given:

• ∠ABE = 24º
• ∠CDE = 66º
• ∠BEC = i
• ∠BAE = j

∠ABC = 66º (opposite angles of parallelogram are equal)

∠ABC = ∠ABE + ∠EBC

66º = 24º + ∠EBC

∴ ∠EBC = 66º – 24º = 42º

∠AEB = ∠EBC (alternate angles, AD || BC)

∴ ∠AEB = 42º

∠DEC = ∠ECB (alternate angles, AD || BC)

But ∠ECB = ∠BEC = i (base angles of an isosceles triangle)

and ∠CED = ∠ECB = i (alternate angles, AD || BC)

Let’s redraw the diagram with our new information.

We can now find i

• ∠AEB = 42º
• ∠BEC = i
• ∠CED = i

∠AEB + ∠BEC + ∠CED = 180º (angles on a straight line)

∴ 42º + i + i = 180º

2i = 180º – 42º

2i = 138º

i = \( \frac{138}{2} \\ = \scriptsize 66º \)

Now let’s find j

We know that

∠ABC + ∠DCB = 180º (consecutive angles of a parallelogram are supplementary)

⇒ ∠ABC = 66º

∠ECB = i = 69º

∠DCB = ∠ECB + ∠ECD

⇒ ∠DCB = 69º + ∠ECD

∴ 66º + 69º + ∠ECD = 180º

∠ECD = 180º – 135º = 45º

∴ ∠DCB = 69º + 45º = 114º

∠DCB = ∠BAE = j (opposite angles of parallelogram are equal)

∴ j = 114º

Cross-check:

We know that the interior angles in a quadrilateral add up to 360º

∠A + ∠B + ∠C + ∠D = 360º

114º + 66º + 114º + 66º = 360º

360º = 360º