In this mini-lesson, we will explore the world of parallelograms and their properties. We will learn about the important theorems related to parallelograms and understand their proofs. So what are we waiting for. Let’s begin!
We all know that a parallelogram is a convex polygon with 4 edges and 4 vertices. The opposite sides are equal and parallel; the opposite angles are also equal.
Let’s play with the simulation given below to better understand a parallelogram and its properties. Try to move the vertices A, B, and D and observe how the figure changes. Observe that at any time, the opposite sides are parallel and equal. Also, the opposite angles are equal. Let’s play along.
Let us dive in and learn more about the parallelograms!
Lesson Plan
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What Are the Properties of a Parallelogram?
It is a type of quadrilateral in which the opposite sides are parallel and equal.
The angles of a parallelogram are the 4 angles formed at the vertices.
- They all add up to 360\(^\circ\)
(\(\angle A +\angle B +\angle C +\angle D = 360^\circ\))
- Opposite angles are equal
(\(\angle A = \angle C\: \text{and} \:\angle B = \angle D\))
- Consecutive angles are supplementary
- \(\angle A +\angle B= 180^\circ\)
- \(\angle B +\angle C= 180^\circ\)
- \(\angle C +\angle D= 180^\circ\)
- \(\angle A +\angle D= 180^\circ\)
Unlock the Mystical World of Parallelograms!
You might be interested in reading these mini lessons for a better understanding of parallelograms.
Explore them and deep dive into the mystical world of parallelograms.
| Click on a mini-lesson to explore further! |
| Parallelograms |
| Quadrilaterals |
| Triangles |
| Pythagoras Theorem |
Properties of a Quadrilateral
Let us first understand the properties of a quadrilateral.
- A quadrilateral is a closed shape with 4 sides.
- All the internal angles of a quadrilateral add up to 360°
Diagonals of a Parallelogram
First, we will recall the meaning of a diagonal.
Diagonals are line segments that join the opposite vertices.
In parallelogram \(PQRS\), \(PR\) and \(QS\) are the diagonals.
- Diagonals of a parallelogram bisect each other
\(OQ =OS\) and \(OR = OP \)
- Each diagonal divides the parallelogram into two congruent triangles.
\(\Delta RSP\cong PQR\)
\(\Delta QPS\cong SRQ\)
- Parallelogram Law: The sum of the squares of the sides is equal to the sum of the squares of the diagonals.
\(PQ^2+QR^2+RS^2+SP^2=QS^2+PR^2\)
Let us explore some theorems based on the properties of a parallelogram.
Theorem 1: Opposite Sides of a Parallelogram Are Equal
In a parallelogram, the opposite sides are equal.
In the figure given below, ABCD is a parallelogram.
\(\therefore\) \(AB=CD\) and \(AD=BC\)
The Converse of Theorem 1
If the opposite sides in a quadrilateral are equal, then it is a parallelogram.
If AB = CD and BC = AD in the given quadrilateral ABCD, then it is a parallelogram.
Unlock the proof of Theorem 1
We will assume that \(ABCD\) is a parallelogram.
Compare \(\Delta ABC\) and \(\Delta CDA\):
\[\begin{align}
& AC=CA \\
&\left( \text{common sides}\right)\\\\
& \angle 1=\angle 4\\
&\left( \text{alternate}\ \text{interior}\ \text{angles} \right)\\\\
& \angle 2=\angle 3 \\
&\left( \text{alternate}\ \text{interior}\ \text{angles} \right)
\end{align}\]
Thus, by the ASA criterion, the two triangles are congruent, which means that the corresponding sides must be equal.
Thus,
\[\begin{align}\boxed{AB=CD\;\text{and}\;AD=BC} \end{align}\]
Unlock the proof of the converse of Theorem 1
Assume that \(ABCD\) is a quadrilateral in which \(AB = CD\) and \(AD = BC\),
Compare \(\Delta ABC\) and \(\Delta CDA\) once again:
\[\begin{align}
& AC=AC\\
&\left( \text{common sides}\right) \\\\
& AB=CD \\
&\left( \text{since alternate interior angles are equal } \right)\\\\
& AD=BC \\
&\left( \text{given}\right)
\end{align}\]
Thus, by the SSS criterion, the two triangles are congruent, which means that the corresponding angles are equal:
\[\begin{align} & \angle 1=\angle 4\Rightarrow AB\parallel CD\ \\ & \angle 2=\angle 3\Rightarrow AD\parallel BC\ \end{align}\]
Hence,
\[\begin{align}\boxed{ AB\parallel CD\;\text{and}\;AD\parallel BC}\end{align}\]
Thus, \(ABCD\) is a parallelogram.
Theorem 2: Opposite Angles of a Parallelogram Are Equal
In a parallelogram, opposite angles are equal.
In the figure given below, ABCD is a parallelogram.
\(\therefore\) \(\angle A=\angle C\) and \(\angle B=\angle D\)
The Converse of Theorem 2
If the opposite angles in a quadrilateral are equal, then it is a parallelogram.
If \(\angle A=\angle C\) and \(\angle B=\angle D\) in the quadrilateral ABCD below, then it is a parallelogram.
Unlock the proof of Theorem 2
First, we assume that \(ABCD\) is a parallelogram.
Compare \(\Delta ABC\) and \(\Delta CDA\) once again:
\[\begin{align}
& AC=AC \\
&\left( \text{common sides}\right) \\\\
& \angle 1=\angle 4 \\
&\left( \text{alternate interior angles} \right) \\\\
& \angle 2=\angle 3 \\
&\left( \text{alternate interior angles} \right)
\end{align}\]
Thus, the two triangles are congruent, which means that
\[\begin{align}\boxed{\angle B=\angle D} \end{align}\]
Similarly, we can show that
\[\begin{align}\boxed{\angle A=\angle C} \end{align}\]
This proves that opposite angles in any parallelogram are equal.
Unlock the proof of the converse of Theorem 2
Assume that \(\angle A\) = \(\angle C\) and \(\angle B\) = \(\angle D\) in the parallelogram ABCD given above.
We have to prove that \(ABCD\) is a parallelogram.
We have:
\[\begin{align}\angle A + \angle B + \angle C + \angle D = \,360^\circ\\2(\angle A + \angle B) =\, 360^\circ\\\angle A + \angle B = \,180^\circ\end{align}\]
This must mean that \(AD\parallel BC\)
Similarly, we can show that \(AB\parallel CD\)
Hence,
\[\begin{align}\boxed{ AD\parallel BC\;\text{and}\;AB\parallel CD}\end{align}\]
and thus, \(ABCD\) is a parallelogram.
Theorem 3: Diagonals of a Parallelogram Bisect Each Other
In a parallelogram, the diagonals bisect each other.
In the figure given below, PQTR is a parallelogram.
PT and QR are the diagonals of PQTR bisecting each other at point E.
\(PE=ET\) and \(ER=EQ\)
The Converse of Theorem 3
If the diagonals in a quadrilateral bisect each other, then it is a parallelogram.
In the quadrilateral PQTR, if PE=ET and ER=EQ, then it is a parallelogram.
Unlock the proof of Theorem 3
First, let us assume that \(PQTR\) is a parallelogram.
Compare \(\Delta RET\) and \(\Delta PEQ\), we have:
\[\begin{align}
& \text{PQ}=\text{RT} \\
&\left( \text{opposite sides of a parallelogram}\right)\\\\
& \angle \text{QRT}=\angle \text{PQR}\\
&\left( \text{alternate interior angles}\right) \\\\
& \angle \text{PTR}=\angle \text{QPT}\\
&\left( \text{alternate interior angles}\right)
\end{align}\]
By the ASA criterion, the two triangles are congruent, which means that
\[\begin{align}\boxed{PE=ET\;\text{and}\;RE=EQ}\end{align}\]
Thus, the two diagonals bisect each other.
Unlock the proof of the converse of Theorem 3
Suppose that the diagonals PT and QR bisect each other.
Compare \(\Delta RET\) and \(\Delta PEQ\) once again. We have:
\[\begin{align} & \text{RE}=\text{EQ} \\
&\left( \text{given}\right) \\\\
& \text{ET}=\text{PE} \\
&\left( \text{given}\right) \\\\
& \angle \text{RET}=\angle \text{PEQ}\\
&\left( \text{vertically opposite angles}\right)
\end{align}\]
By the SAS criterion, the two triangles are congruent, which means that:
\(\angle \text{QRT}\) = \(\angle \text{PQR}\)
\(\angle \text{PTR}\) = \(\angle \text{QPT}\)
Hence,
\[\begin{align}\boxed{PQ\parallel RT\;{\rm{and}}\;PR\parallel QT} \end{align}\]
Thus, \(PQRT\) is a parallelogram.
Theorem 4: One Pair of Opposite Sides Is Equal and Parallel in a Parallelogram
It is given that \(AB=CD\) \(\)and \(AB || CD \) in the above figure.
We can prove that \(ABCD\) is a parallelogram.
Unlock the proof of Theorem 4
Compare \(\Delta AEB\) and \(\Delta DEC\). We have:
\[\begin{align}
& AB=CD\\
&\left( \text{given}\right)\\\\
& \angle 1=\angle 3 \\
&\left( \text{alternate interior angles}\right)\\\\
& \angle 2=\angle 4\\
&\left( \text{alternate interior angles}\right)
\end{align}\]
Thus, the two triangles are congruent, which means that
\[\begin{align}\boxed{AE=EC\;\text{and}\;BE=ED}\end{align}\]
Therefore, the diagonals AC and BD bisect each other, and this further means that \(ABCD\) is a parallelogram.
|
1. We have shown that the following statements are equivalent, that is, you can use them interchangeably. A quadrilateral is a parallelogram when: |
2. Note that the relation between two lines intersected by a transversal, when the angles on the same side of the transversal are supplementary, are parallel to each other. |
Solved Examples
| Example 1 |
If one angle of a parallelogram is 90o, show that all its angles will be equal to 90o.
Solution:
Consider the parallelogram \(ABCD\) in the following figure, in which \(\angle A\) is a right angle:
We know that in any parallelogram, the opposite angles are equal.
This implies, \(\angle C\) must be 90o
Also, in any parallelogram, the adjacent angles are supplementary.
This implies \(\angle B=180^\circ - \angle A\)
Similarly, \(\angle D=180^\circ - \angle C\)
\(\begin{align}\angle B = \angle D &=\,180^\circ - \;90^\circ \\\\&=\,90^\circ\end{align}\)
Hence,
\[\begin{align}\boxed{\angle A=\angle B=\angle C=\angle D = 90^\circ} \end{align}\]
Clearly, all the angles in this parallelogram (which is actually a rectangle) are equal to 90o.
| \(\therefore\) when one angle of a parallelogram is 90o, the parallelogram is a rectangle |
| Example 2 |
What is the difference between the opposite angles of a parallelogram?
Solution:
In a parallelogram, the opposite sides and opposite angles are equal.
Therefore, the difference between the opposite angles of a parallelogram is:
\(x-x = 0^\circ \)
\(y-y = 0^\circ \)
| \(\therefore\) Difference between opposite angles of a parallelogram is 0° |
| Example 3 |
In a quadrilateral \(ABCD\), the diagonals \(AC\) and \(BD\) bisect each other at right angles.
Show that the quadrilateral is a rhombus.
Solution:
Consider the following figure:
First of all, we note that since the diagonals bisect each other, we can conclude that \(ABCD\) is a parallelogram.
Now, let us compare \(\Delta AEB\) and \(\Delta AED\):
\[\begin{align} AE&=AE \left( \text{common}\right) \\\\ BE&=ED \left( \text{given}\right) \\\\ \angle AEB&=\angle AED=\,90^\circ \left( \text{given}\right) \end{align}\]
Thus, by the SAS criterion, the two triangles are congruent, which means that
\(AB\) = \(CD\)
This further means that
\[\begin{align}\boxed{ AB=BC=CD=AD} \end{align}\]
Clearly, \(ABCD\) is a rhombus.
| \(\therefore\) Parallelogram ABCD is a rhombus |
| Example 4 |
\(ABCD\) is a quadrilateral in which the diagonals bisect each other.
Show that \(B\) and \(D\) are equidistant from \(AC\).
Solution:
Since its diagonals bisect each other, \(ABCD\) is a parallelogram.
Consider the following figure:
Essentially, we have to show that
\(BF\) = \(DE\)
Compare \(\Delta BFG\) with \(\Delta DEG\).
\[\begin{align} & BG=GD\ \ \ \ \\&\left( \text{diagonals bisect each other}\right) \\\\ & \angle BGF=\angle DGE\ \ \ \ \ \ \\&\left( \text{vertically opposite angles}\right) \\\\ & \angle 1=\angle 2\ \ \ \ \ \ \\&\left( \text{alternate interior angles}\right) \end{align}\]
By the ASA criterion, the two triangles are congruent, which means that:
\[\begin{align}\boxed{ BF=DE} \end{align}\]
Thus, \(B\) and \(D\) are equidistant from \(A\).
Note: Two lines that are perpendicular to the same line are parallel to each other.
| \(\therefore\) B and D are equidistant from AC |
| Example 5 |
Prove that the bisectors of the angles in a parallelogram form a rectangle.
Solution:
Consider the following figure, in which \(ABCD\) is a parallelogram, and the dotted lines represent the (four) angle bisectors.
We have to show that \(EFGH\) is a rectangle:
We can show this by proving that each of the four angles of \(EFGH\) is a right angle.
In parallelogram \(ABCD\), we have:
\(\angle A + \angle B = 180^\circ \)
Now, consider \(\Delta AHB\).
We have:
\(\begin{align}\angle 1 + \angle 2 =& \frac{1}{2}\left( {\angle A + \angle B} \right)\\\\ =&\,\ 90^\circ\end{align}\)
Thus,
\[\begin{align}\boxed{\angle 3 = 90^\circ} \end{align}\]
Similarly, we can prove that each of the other three angles of quadrilateral \(EFGH\) is a right angle.
This means that \(EFGH\) is a rectangle.
| \(\therefore\) Bisectors of the angles in a parallelogram form a rectangle |
- Why is a kite not a parallelogram?
- Is an isosceles trapezoid a parallelogram?
Interactive Questions on Properties of Parallelograms
Here are a few problems for you to practice.
Select/Type your answer and click the "Check Answer" button to see the result.
- If the opposite sides of a quadrilateral are equal and parallel, then it is a parallelogram.
- If the opposite angles in a quadrilateral are equal, then it is a parallelogram.
- A parallelogram that has all equal sides is a rhombus.
Let's summarize:
The mini-lesson was aimed at helping you learn about parallelograms and their properties. Hope you enjoyed learning about them and exploring the important theorems related to parallelograms.
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Frequently Asked Questions(FAQs)
1. What are the 7 properties of a parallelogram?
The 7 properties of a parallelogram are:
- The opposite sides of a parallelogram are equal.
- If the opposite sides of a quadrilateral are equal, it is a parallelogram.
- The opposite angles of a parallelogram are equal.
- If the opposite angles of a quadrilateral are equal, it is a parallelogram.
- The diagonals of a parallelogram bisect each other.
- If the diagonals of a quadrilateral bisect each other, it is a parallelogram.
- If one pair of opposite sides of a quadrilateral is equal and parallel, then the quadrilateral is a parallelogram.
2. What are the properties of the diagonals of a parallelogram?
The properties of the diagonals of a parallelogram are:
- A diagonal of a parallelogram divides it into two congruent triangles.
- The diagonals of a parallelogram bisect each other.