Quadrilaterals

Properties of a Parallelogram

You have already studied quadrilaterals and their types in Class VIII. A quadrilateral has four sides, four angles and four vertices. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

Let us perform an activity.

 Fig. 8.1

      Cut out a parallelogram from a sheet of paper and cut it along a diagonal (see Fig. 8.1). You obtain two triangles. What can you say about these triangles?

      Place one triangle over the other. Turn one around, if necessary. What do you observe?

      Observe that the two triangles are congruent to each other. Repeat this activity with some more parallelograms.

      Each time you will observe that each diagonal divides the parallelogram into two congruent triangles.

      Let us now prove this result.

Theorem 8.1 : A diagonal of a parallelogram divides it into two congruent triangles.

Proof : Let ABCD be a parallelogram and AC be a diagonal (see Fig. 8.2). Observe that the diagonal AC divides parallelogram ABCD into two triangles, namely, ∆ ABC and ∆ CDA. We need to prove that these triangles are congruent.

In ∆ ABC and ∆ CDA, note that BC || AD and AC is a transversal.

 Fig. 8.2

So,     ∠ BCA = ∠ DAC (Pair of alternate angles)

Also, AB || DC and AC is a transversal.

So,     ∠ BAC = ∠ DCA (Pair of alternate angles)

and AC = CA (Common)

So,     ∆ ABC ≅ ∆ CDA (ASA rule)

or, diagonal AC divides parallelogram ABCD into two congruent triangles ABC and CDA.

Theorem 8.2 : In a parallelogram, opposite sides are equal.

      You have already proved that a diagonal divides the parallelogram into two congruent triangles; so what can you say about the corresponding parts say, the corresponding sides? They are equal.

So, AB = DC and AD = BC

      Now what is the converse of this result? You already know that whatever is given in a theorem, the same is to be proved in the converse and whatever is proved in the theorem it is given in the converse. Thus, Theorem 8.2 can be stated as given below :

If a quadrilateral is a parallelogram, then each pair of its opposite sides is equal. So its converse is :

Theorem 8.3 : If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Can you reason out why?

 Fig. 8.3

Let sides AB and CD of the quadrilateral ABCD be equal and also AD = BC (see Fig. 8.3). Draw diagonal AC.

Clearly,             ∆ ABC ≅ ∆ CDA (Why?)

So,                  ∠ BAC = ∠ DCA

and                 ∠ BCA = ∠ DAC (Why?)

Can you now say that ABCD is a parallelogram? Why?

You have just seen that in a parallelogram each pair of opposite sides is equal and conversely if each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. Can we conclude the same result for the pairs of opposite angles?

            Draw a parallelogram and measure its angles. What do you observe?

            Each pair of opposite angles is equal.

            Repeat this with some more parallelograms. We arrive at yet another result as given below.

Theorem 8.4 : In a parallelogram, opposite angles are equal.

        Now, is the converse of this result also true? Yes. Using the angle sum property of a quadrilateral and the results of parallel lines intersected by a transversal, we can see that the converse is also true. So, we have the following theorem :

Theorem 8.5 : If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

         There is yet another property of a parallelogram. Let us study the same. Draw a parallelogram ABCD and draw both its diagonals intersecting at the point O (see Fig. 8.4).

 Fig. 8.4

Measure the lengths of OA, OB, OC and OD.

What do you observe? You will observe that

             OA = OC and OB = OD.

or, O is the mid-point of both the diagonals.

Repeat this activity with some more parallelograms.

Each time you will find that O is the mid-point of both the diagonals.

So, we have the following theorem :

Theorem 8.6 : The diagonals of a parallelogram bisect each other.

             Now, what would happen, if in a quadrilateral the diagonals bisect each other? Will it be aparallelogram? Indeed this is true.

            This result is the converse of the result of Theorem 8.6. It is given below:

Theorem 8.7 : If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

You can reason out this result as follows:

 Fig. 8.5

Note that in Fig. 8.5, it is given that OA = OC and OB = OD.

So,           ∆ AOB ≅ ∆ COD (Why?)

Therefore, ∠ ABO = ∠ CDO (Why?)

From this, we get AB || CD

Similarly, BC || AD Therefore ABCD is a parallelogram.

Let us now take some examples.

Example 1 : Show that each angle of a rectangle is a right angle.

Solution : Let us recall what a rectangle is.

A rectangle is a parallelogram in which one angle is a right angle.

Let ABCD be a rectangle in which ∠ A = 90°.

We have to show that ∠ B = ∠ C = ∠ D = 90°

We have, AD || BC and AB is a transversal (see Fig. 8.6).

 Fig. 8.6

So,      ∠ A + ∠ B = 180° (Interior angles on the same side of the transversal)

But,     ∠ A = 90° So,      ∠ B = 180° – ∠ A = 180° – 90° = 90°

Now,   ∠ C = ∠ A and ∠ D = ∠ B (Opposite angles of the parallellogram)

So,      ∠ C = 90° and ∠ D = 90°.

Therefore, each of the angles of a rectangle is a right angle.

Example 2 : Show that the diagonals of a rhombus are perpendicular to each other.

Solution : Consider the rhombus ABCD (see Fig. 8.7).

 Fig. 8.7

You know that AB = BC = CD = DA (Why?)

Now, in ∆ AOD and ∆ COD, OA = OC (Diagonals of a parallelogram bisect each other)

OD = OD            (Common)

AD = CD

Therefore, ∆ AOD ≅ ∆ COD (SSS congruence rule)

This gives, ∠ AOD = ∠ COD (CPCT)

But, ∠ AOD + ∠ COD = 180° (Linear pair)

So, 2∠ AOD = 180°

or, ∠ AOD = 90°

So, the diagonals of a rhombus are perpendicular to each other

Example 3 : ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || AB (see Fig. 8.8). Show that

(i) ∠ DAC = ∠ BCA and (ii) ABCD is a parallelogram.

Solution : (i) ∆ ABC is isosceles in which AB = AC (Given)

So,        ∠ ABC = ∠ ACB (Angles opposite to equal sides)

Also,      ∠ PAC = ∠ ABC + ∠ ACB (Exterior angle of a triangle)

or, ∠ PAC = 2∠ ACB                                                (1)

Now, AD bisects ∠ PAC.

So, ∠ PAC = 2∠ DAC                                               (2)

 Fig. 8.8

Therefore,

      2∠ DAC = 2∠ ACB [From (1) and (2)]

or, ∠ DAC = ∠ ACB

(ii) Now, these equal angles form a pair of alternate angles when line segments BC and AD are intersected by a transversal AC.

So, BC || AD

Also, BA || CD (Given)

Now, both pairs of opposite sides of quadrilateral ABCD are parallel.

So, ABCD is a parallelogram

Example 4 : Two parallel lines l and m are intersected by a transversal p (see Fig. 8.9). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

Solution : It is given that PS || QR and transversal p intersects them at points A and C respectively. The bisectors of ∠ PAC and ∠ ACQ intersect at B and bisectors of ∠ ACR and ∠ SAC intersect at D. We are to show that quadrilateral ABCD is a rectangle.

 Fig. 8.9

Now, ∠ PAC = ∠ ACR (Alternate angles as l || m and p is a transversal)

So, \(\frac{1}{2}\) ∠ PAC = \(\frac{1}{2}\) ∠ ACR

i.e., ∠ BAC = ∠ ACD

These form a pair of alternate angles for lines AB and DC with AC as transversal and they are equal also.

So, AB || DC

Similarly, BC || AD (Considering ∠ ACB and ∠ CAD)

Therefore, quadrilateral ABCD is a parallelogram.

Also, ∠ PAC + ∠ CAS = 180° (Linear pair)

So, \(\frac{1}{2}\) ∠ PAC + \(\frac{1}{2}\) ∠ CAS = \(\frac{1}{2}\) × 180° = 90°

or, ∠ BAC + ∠ CAD = 90°

or, ∠ BAD = 90°

So, ABCD is a parallelogram in which one angle is 90°.

Therefore, ABCD is a rectangle.

Example 5 : Show that the bisectors of angles of a parallelogram form a rectangle.

Solution : Let P, Q, R and S be the points of intersection of the bisectors of ∠ A and ∠ B, ∠ B and ∠ C, ∠ C and ∠ D, and ∠ D and ∠ A respectively of parallelogram ABCD (see Fig. 8.10). In ∆ ASD, what do you observe? Since DS bisects ∠ D and AS bisects ∠ A, therefore,

 Fig. 8.10

∠ DAS + ∠ ADS = \(\frac{1}{2}\) ∠ A + \(\frac{1}{2}\) ∠ D

                       =\(\frac{1}{2}\) (∠ A + ∠ D)

                       =\(\frac{1}{2}\) × 180° (∠ A and ∠ D are interior angles on the same side of the transversal)

                       = 90°

Also, ∠ DAS + ∠ ADS + ∠ DSA = 180° (Angle sum property of a triangle)

or,                      90° + ∠ DSA = 180°

or,                                ∠ DSA = 90°

So,                               ∠ PSR = 90° (Being vertically opposite to ∠ DSA)

Similarly, it can be shown that ∠ APB = 90° or ∠ SPQ = 90° (as it was shown for ∠ DSA). Similarly, ∠ PQR = 90° and ∠ SRQ = 90°.

So, PQRS is a quadrilateral in which all angles are right angles.

        Can we conclude that it is a rectangle? Let us examine. We have shown that ∠ PSR = ∠ PQR = 90° and ∠ SPQ = ∠ SRQ = 90°. So both pairs of opposite angles are equal.

Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90° and so, PQRS is a rectangle.

Quadrilaterals