Properties of Parallelograms | Mathematics (Maths) Class 9 PDF Download

Introduction

It is a four-sided closed figure with opposite sides and equal angles. 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 learn more about the properties of parallelograms in detail in this lesson.

What are the Properties of a Parallelogram?

The four important properties of a parallelogram are as follows.

• Opposites sides of a parallelogram are equal and parallel to each other.
• Opposite angles are equal. ∠A= ∠C, and ∠B = ∠D
• All the angles of a parallelogram add up to 360^o.  ∠A + ∠B + ∠C + ∠D = 360^o. 
• The consecutive angles of a parallelogram are supplementary
• ∠A + ∠B  = 180^o
• ∠B + ∠C  = 180^o
• ∠C + ∠D = 180^o
• ∠D + ∠A = 180^o 

Sides and Angles of a Parallelogram 

Theorems on Properties of a Parallelogram

• Opposite sides of a parallelogram are equal
• Opposite angles of a parallelogram are equal
• Diagonals of a parallelogram bisect each other
• One pair of opposite sides is equal and parallel in a parallelogram

Theorem 1: A diagonal of a parallelogram divides it into two congruent triangles.

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. The properties of diagonals of a parallelogram are as follows:

• Diagonals of a parallelogram bisect each other. OQ =OS  and OR = OP 
• Each diagonal divides the parallelogram into two congruent triangles, so, ΔRSP ≅ ΔPQR and ΔQPS ≅ ΔSRQ.
• Parallelogram Law: The sum of the squares of the sides is equal to the sum of the squares of the diagonals. PQ²+QR²+RS²+SP² = QS²+PR²

Diagonals of a Parallelogram 

Theorem 2: In a Parallelogram the Opposite Sides Are Equal. This means, in a parallelogram, the opposite sides are equal.
Given: ABCD is a parallelogram.
To Prove: The opposite sides are equal, AB=CD, and BC=AD. 

Proof: In parallelogram ABCB, compare triangles ABC and CDA. In these triangles AC = CA (common sides). Also ∠BAC =∠DCA (alternate interior angles), and ∠BCA = ∠DAC (alternate interior angles). Hence by the ASA criterion, both the triangles are congruent and the corresponding sides are equal.  Therefore we have AB = CD, and BC = AD.

Theorem 3: 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.
Given: The opposite sides in a quadrilateral ABCD are equal, AB = CD, and BC = AD.
To Prove: ABCD is a parallelogram.
Proof: In the quadrilateral ABCD we are given that AB = CD, and AD = BC. Now compare the two triangles ABC, and CDA. Here we have AC = AC (Common sides), AB = CD (since alternate interior angles are equal), and AD = BC (given). Thus by the SSS criterion both the triangles are congruent, and the corresponding angles are equal. Hence we can conclude that ∠BAC = ∠DCA, and ∠BCA = ∠DAC.  Therefore AB // CD, BC // AD, and ABCD is a parallelogram.

Theorem 4: In a Parallelogram, the Opposite Angles Are Equal. 
Given: ABCD is a parallelogram, and ∠A, ∠B,  ∠C, ∠D are the four angles.
To Prove: ∠A =∠C and ∠B=∠D

Proof: Let us assume that ABCD is a parallelogram. 
Now compare triangles ABC, and CDA. 
Here we have  AC=AC (common sides),  ∠1=∠4 (alternate interior angles), and ∠2=∠3 (alternate interior angles). 
Thus, the two triangles are congruent, which means that ∠B=∠D. Similarly, we can show that ∠A=∠C. 
This proves that opposite angles in any parallelogram are equal.

Theorem 5: If the opposite angles in a quadrilateral are equal, then it is a parallelogram.
Given: ∠A=∠C and ∠B=∠D in the quadrilateral ABCD. To Prove: ABCD is a parallelogram.
Proof: Assume that ∠A = ∠C and ∠B = ∠D in the parallelogram ABCD given above. 
We have to prove that ABCD is a parallelogram. 
We have: ∠A + ∠B + ∠C + ∠D = 360º;2(∠A + ∠B) =360º; ∠A + ∠B = 180º. 
This must mean that AD // BC. Similarly, we can show that AB//CD. Hence, AD//BC, and AB//CD. 
Therefore ABCD is a parallelogram.

Theorem 6: Diagonals of a Parallelogram Bisect Each Other. That means, in a parallelogram, the diagonals bisect each other. 
Given: PQTR is a parallelogram. PT and QR are the diagonals of the parallelogram.
To Prove: The diagonals PT, and RQ bisect each other. PE=ET and ER=EQ

Proof: First, let us assume that PQTR is a parallelogram. Compare triangles RET, and triangle PEQ. 
We have PQ = RT (opposite sides of the parallelogram), ∠QRT = ∠PQR (alternate interior angles), and ∠PTR = ∠QPT  (alternate interior angles). 
By the ASA criterion, the two triangles are congruent, which means that PE = ET, and RE = EQ. 
Thus, the two diagonals PT, and RQ bisect each other, and PE=ET and ER=EQ

Theorem 7: 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.
Given: The diagonals  PT and QR bisect each other.
To Prove: PQRT is a parallelogram.
Proof: Suppose that the diagonals PT and QR bisect each other. Compare triangle RET, and triangle PEQ once again. 
We have: RE = EQ, ET = PE (Diagonals bisect each other), ∠RET =∠PEQ  (vertically opposite angles). 
Hence by the SAS criterion, the two triangles are congruent. 
This means that ∠QRT = ∠PQR, and ∠PRT = ∠QPT . 
Hence, PQ//RT, and RT//QT. 
Thus PQRT is a parallelogram.

Theorem 8: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Proof: You can prove this theorem using the following clue:

Observe above figure in which E and F are mid-points of AB and AC respectively and CD || BA.
∆ AEF ≅ ∆ CDF (ASA Rule)
So, EF = DF and BE = AE = DC (Why?)
Therefore, BCDE is a parallelogram. (Why?)
This gives EF || BC.
In this case, also note that EF = 1/2ED = 1/2BC.
Can you state the converse of Theorem 8.8? Is the converse true?
You will see that converse of the above theorem is also true which is stated as below:

Theorem 9: The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

In above figure, observe that E is the mid-point of AB, line l is passing through E and is parallel to BC and CM || BA.

Solved Examples

Sol: A rectangle is a parallelogram in which each angle is 90° and parallelogram on the same base and between same parallels are equal in area.

∴ ar. (||gmABCD) = ar. (rectangle ABPQ)

But ar. (||gmABPQ) = AB×AQ

= (8 × 6) cm² = 48 cm²

∴ Area of parallelogram ABCD = 48 cm².

Example 2: In a parallelogram ABCD, AB = 8 cm. The altitudes corresponding to sides AB and AD are respectively 4 m and 5 cm. Find AD.

Sol: We know that the area of a parallelogram = Base × Corresponding altitude.

∴ Area of parallelogram ABCD = AD × BN = AB × DM

⇒ AD × 5 = 8 × 4

⇒ AD =

= 6.4 cm. 

Example 3: ABCD is a quadrilateral and BD is one its diagonals as shown in figure. Show that ABCD is a parallelogram and find its area.

Sol: In ΔABD and ΔCDB

∠ABD = ∠CBD = 90o (alternate interior angles are equal)

∠ADB = ∠CBD

∠BDA = ∠BDC

Hence, ABCD is a parallelogram.

Now, area of parallelogram ABCD = (base ×corresponding altitude)

⇒area of parallelogram ABCD = AB × BD = 5 ×12 sq. units = 60 sq. cms.

Example 4: Prove that parallelograms on the same base and between the same parallel lines are equal in area.

Sol: Given: Two parallelogram ABCD and PBCQ having same base BC and between the same parallel lines BC and AQ.

To Prove: ar(parallelogram ABCD)= ar (parallelogram PBCQ)

Proof: In triangles ABP and DCQ,

∠BAP = ∠CDQ

(Corresponding angles when AQ intersects parallel lines AB and DC).

∠BPA = ∠CDQ

(Corresponding angles when AQ intersects parallel lines BP and CQ).

AB = DC (Opposite sides of a parallelogram)

∴ΔABP ≅ ΔDCQ (AAS)

Hence area ΔABP = area ΔDCQ 

or, area ΔABP + area BPDC = area ΔDCQ + area Δ BPDC

∴ar (parallelogram ABCD) = ar (parallelogram PBCQ).