CHAPTER 8

QUADRILATERALS

8.1 IntroductionYou have studied many properties of a triangle in Chapters 6 and 7 and you know thaton joining three non-collinear points in pairs, the figure so obtained is a triangle. Now,let us mark four points and see what we obtain on joining them in pairs in some order.

Fig. 8.1

Note that if all the points are collinear (in the same line), we obtain a linesegment [see Fig. 8.1 (i)], if three out of four points are collinear, we get a triangle[see Fig. 8.1 (ii)], and if no three points out of four are collinear, we obtain a closedfigure with four sides [see Fig. 8.1 (iii) and (iv)].

Such a figure formed by joining four points in an order is called a quadrilateral.In this book, we will consider only quadrilaterals of the type given in Fig. 8.1 (iii) butnot as given in Fig. 8.1 (iv).

A quadrilateral has four sides, four angles and four vertices [see Fig. 8.2 (i)].

Fig. 8.2

136 MATHEMATICS

In quadrilateral ABCD, AB, BC, CD and DA are the four sides; A, B, C and D arethe four vertices and ∠ A, ∠ B, ∠ C and ∠ D are the four angles formed at thevertices.

Now join the opposite vertices A to C and B to D [see Fig. 8.2 (ii)].AC and BD are the two diagonals of the quadrilateral ABCD.In this chapter, we will study more about different types of quadrilaterals, their

properties and especially those of parallelograms.You may wonder why should we study about quadrilaterals (or parallelograms)

Look around you and you will find so many objects which are of the shape of aquadrilateral - the floor, walls, ceiling, windows of your classroom, the blackboard,each face of the duster, each page of your book, the top of your study table etc. Someof these are given below (see Fig. 8.3).

Fig. 8.3

Although most of the objects we see around are of the shape of special quadrilateralcalled rectangle, we shall study more about quadrilaterals and especially parallelogramsbecause a rectangle is also a parallelogram and all properties of a parallelogram aretrue for a rectangle as well.

8.2 Angle Sum Property of a QuadrilateralLet us now recall the angle sum property of aquadrilateral.

The sum of the angles of a quadrilateral is 360º.This can be verified by drawing a diagonal and dividingthe quadrilateral into two triangles.

Let ABCD be a quadrilateral and AC be adiagonal (see Fig. 8.4).

What is the sum of angles in ∆ ADC? Fig. 8.4

QUADRILATERALS 137

You know that∠ DAC + ∠ ACD + ∠ D = 180° (1)

Similarly, in ∆ ABC,∠ CAB + ∠ ACB + ∠ B = 180° (2)

Adding (1) and (2), we get∠ DAC + ∠ ACD + ∠ D + ∠ CAB + ∠ ACB + ∠ B = 180° + 180° = 360°

Also, ∠ DAC + ∠ CAB = ∠ A and ∠ ACD + ∠ ACB = ∠ CSo, ∠ A + ∠ D + ∠ B + ∠ C = 360°.i.e., the sum of the angles of a quadrilateral is 360°.

8.3 Types of QuadrilateralsLook at the different quadrilaterals drawn below:

Fig. 8.5

Observe that :One pair of opposite sides of quadrilateral ABCD in Fig. 8.5 (i) namely, ABand CD are parallel. You know that it is called a trapezium.Both pairs of opposite sides of quadrilaterals given in Fig. 8.5 (ii), (iii) , (iv)and (v) are parallel. Recall that such quadrilaterals are called parallelograms.So, quadrilateral PQRS of Fig. 8.5 (ii) is a parallelogram.

138 MATHEMATICS

Similarly, all quadrilaterals given in Fig. 8.5 (iii), (iv) and (v) are parallelograms.In parallelogram MNRS of Fig. 8.5 (iii), note that one of its angles namely∠ M is a right angle. What is this special parallelogram called? Try to recall.It is called a rectangle.The parallelogram DEFG of Fig. 8.5 (iv) has all sides equal and we know thatit is called a rhombus.The parallelogram ABCD of Fig. 8.5 (v) has ∠ A = 90° and all sides equal; itis called a square.In quadrilateral ABCD of Fig. 8.5 (vi), AD = CD and AB = CB i.e., two pairsof adjacent sides are equal. It is not a parallelogram. It is called a kite.

Note that a square, rectangle and rhombus are all parallelograms.A square is a rectangle and also a rhombus.A parallelogram is a trapezium.A kite is not a parallelogram.A trapezium is not a parallelogram (as only one pair of opposite sides is parallelin a trapezium and we require both pairs to be parallel in a parallelogram).A rectangle or a rhombus is not a square.

Look at the Fig. 8.6. We have a rectangle and a parallelogram with same perimeter14 cm.

Fig. 8.6

Here the area of the parallelogram is DP × AB and this is less than the area of therectangle i.e., AB × AD as DP < AD. Generally sweet shopkeepers cut ‘Burfis’ in theshape of a parallelogram to accomodate more pieces in the same tray (see the shapeof the Burfi before you eat it next time!).

Let us now review some properties of a parallelogram learnt in earlier classes.

QUADRILATERALS 139

8.4 Properties of a Parallelogram

Let us perform an activity.Cut out a parallelogram from a sheet of paper

and cut it along a diagonal (see Fig. 8.7). You obtaintwo triangles. What can you say about thesetriangles?

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

Observe that the two triangles are congruent toeach other.

Repeat this activity with some more parallelograms. Each time you will observethat 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 congruenttriangles.Proof : Let ABCD be a parallelogram and AC be a diagonal (see Fig. 8.8). Observethat the diagonal AC divides parallelogram ABCD into two triangles, namely, ∆ ABCand ∆ CDA. We need to prove that these triangles are congruent.

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

Now, measure the opposite sides of parallelogram ABCD. What do you observe?You will find that AB = DC and AD = BC.This is another property of a parallelogram stated below:

Theorem 8.2 : In a parallelogram, opposite sides are equal.You have already proved that a diagonal divides the parallelogram into two congruent

Fig. 8.7

Fig. 8.8

140 MATHEMATICS

triangles; so what can you say about the corresponding parts say, the correspondingsides? They are equal.So, AB = DC and AD = BC

Now what is the converse of this result? You already know that whatever is givenin a theorem, the same is to be proved in the converse and whatever is proved in thetheorem 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. Soits converse is :

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

Can you reason out why?Let sides AB and CD of the quadrilateral ABCD

be equal and also AD = BC (see Fig. 8.9). Drawdiagonal AC.

Clearly, ∆ ABC ≅ ∆ CDA (Why?)So, ∠ BAC = ∠ DCAand ∠ 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 aparallelogram. 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 seethat 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, thenit is a parallelogram.

Fig. 8.9

QUADRILATERALS 141

There is yet another property of a parallelogram. Let us study the same. Draw aparallelogram ABCD and draw both its diagonals intersecting at the point O(see Fig. 8.10).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 parallelogrambisect each other.

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

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

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

You can reason out this result as follows:Note that in Fig. 8.11, it is given that OA = OC

and OB = OD.So, ∆ AOB ≅ ∆ COD (Why?)Therefore, ∠ ABO = ∠ CDO (Why?)From this, we get AB || CDSimilarly, BC || ADTherefore 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

Fig. 8.10

Fig. 8.11

142 MATHEMATICS

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. 8.12).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. 8.13).You know that AB = BC = CD = DA (Why?)Now, in ∆ AOD and ∆ COD,

OA = OC (Diagonals of a parallelogrambisect 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

Fig. 8.12

Fig. 8.13

QUADRILATERALS 143

angle PAC and CD || AB (see Fig. 8.14). 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)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 BCand AD are intersected by a transversal AC.So, BC || ADAlso, 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. 8.15). Show that the quadrilateral formed by the bisectors of interior anglesis a rectangle.Solution : It is given that PS || QR and transversal p intersects them at points A andC 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 arectangle.Now, ∠ PAC = ∠ ACR(Alternate angles as l || m and p is a transversal)

So,12 ∠ PAC =

12 ∠ ACR

i.e., ∠ BAC = ∠ ACD

Fig. 8.14

Fig. 8.15

144 MATHEMATICS

These form a pair of alternate angles for lines AB and DC with AC as transversal andthey are equal also.So, AB || DCSimilarly, BC || AD (Considering ∠ ACB and ∠ CAD)Therefore, quadrilateral ABCD is a parallelogram.Also, ∠ PAC + ∠ CAS = 180° (Linear pair)

So,12

∠ PAC +12

∠ CAS =12

× 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 ofintersection of the bisectors of ∠ A and ∠ B, ∠ Band ∠ C, ∠ C and ∠ D, and ∠ D and ∠ A respectivelyof parallelogram ABCD (see Fig. 8.16).In ∆ ASD, what do you observe?Since DS bisects ∠ D and AS bisects ∠ A, therefore,

∠ DAS + ∠ ADS =12

∠ A + 12

∠ D

=12 (∠ A + ∠ D)

=12

× 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)

Fig. 8.16

QUADRILATERALS 145

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 anglesare equal.Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90° andso, PQRS is a rectangle.

8.5 Another Condition for a Quadrilateral to be a ParallelogramYou have studied many properties of a parallelogram in this chapter and you have alsoverified that if in a quadrilateral any one of those properties is satisfied, then it becomesa parallelogram.

We now study yet another condition which is the least required condition for aquadrilateral to be a parallelogram.

It is stated in the form of a theorem as given below:

Theorem 8.8 : A quadrilateral is a parallelogram if a pair of opposite sides isequal and parallel.

Look at Fig 8.17 in which AB = CD andAB || CD. Let us draw a diagonal AC. You can showthat ∆ ABC ≅ ∆ CDA by SAS congruence rule.

So, BC || AD (Why?)Let us now take an example to apply this property

of a parallelogram.

Example 6 : ABCD is a parallelogram in which Pand Q are mid-points of opposite sides AB and CD(see Fig. 8.18). If AQ intersects DP at S and BQintersects CP at R, show that:(i) APCQ is a parallelogram.(ii) DPBQ is a parallelogram.(iii) PSQR is a parallelogram.

Fig. 8.17

Fig. 8.18

146 MATHEMATICS

Solution : (i) In quadrilateral APCQ,AP || QC (Since AB || CD) (1)

AP =12

AB, CQ = 12

CD (Given)

Also, AB = CD (Why?)So, AP = QC (2)Therefore, APCQ is a parallelogram [From (1) and (2) and Theorem 8.8](ii) Similarly, quadrilateral DPBQ is a parallelogram, because

DQ || PB and DQ = PB(iii) In quadrilateral PSQR,

SP || QR (SP is a part of DP and QR is a part of QB)Similarly, SQ || PRSo, PSQR is a parallelogram.

EXERCISE 8.11. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the

quadrilateral.

2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then itis a rhombus.

4. Show that the diagonals of a square are equal and bisect each other at right angles.

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at rightangles, then it is a square.

6. Diagonal AC of a parallelogram ABCD bisects∠ A (see Fig. 8.19). Show that

(i) it bisects ∠ C also,

(ii) ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal ACbisects ∠ A as well as ∠ C and diagonal BDbisects ∠ B as well as ∠ D.

8. ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that:(i) ABCD is a square (ii) diagonal BD bisects ∠ B as well as ∠ D.

Fig. 8.19

QUADRILATERALS 147

9. In parallelogram ABCD, two points P and Q aretaken on diagonal BD such that DP = BQ(see Fig. 8.20). Show that:

(i) ∆ APD ≅ ∆ CQB

(ii) AP = CQ(iii) ∆ AQB ≅ ∆ CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

10. ABCD is a parallelogram and AP and CQ areperpendiculars from vertices A and C on diagonalBD (see Fig. 8.21). Show that

(i) ∆ APB ≅ ∆ CQD

(ii) AP = CQ

11. In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EFand BC || EF. Vertices A, B and C are joined tovertices D, E and F respectively (see Fig. 8.22).Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD || CF and AD = CF(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ∆ ABC ≅ ∆ DEF.

12. ABCD is a trapezium in which AB || CD andAD = BC (see Fig. 8.23). Show that

(i) ∠ A = ∠ B

(ii) ∠ C = ∠ D

(iii) ∆ ABC ≅ ∆ BAD

(iv) diagonal AC = diagonal BD

[Hint : Extend AB and draw a line through Cparallel to DA intersecting AB produced at E.]

Fig. 8.20

Fig. 8.21

Fig. 8.22

Fig. 8.23

148 MATHEMATICS

8.6 The Mid-point Theorem

You have studied many properties of a triangle as well as a quadrilateral. Now let usstudy yet another result which is related to the mid-point of sides of a triangle. Performthe following activity.

Draw a triangle and mark the mid-points E and F of two sides of the triangle. Jointhe points E and F (see Fig. 8.24).

Measure EF and BC. Measure ∠ AEF and ∠ ABC.What do you observe? You will find that :

EF = 12

BC and ∠ AEF = ∠ ABC

so, EF || BCRepeat this activity with some more triangles.So, you arrive at the following theorem:

Theorem 8.9 : The line segment joining the mid-points of two sides of a triangleis parallel to the third side.

You can prove this theorem using the followingclue:

Observe Fig 8.25 in which E and F are mid-pointsof 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 = 12 ED =

12 BC.

Can you state the converse of Theorem 8.9? Is the converse true?You will see that converse of the above theorem is also true which is stated as

below:

Theorem 8.10 : The line drawn through the mid-point of one side of a triangle,parallel to another side bisects the third side.

Fig. 8.25

Fig. 8.24

QUADRILATERALS 149

In Fig 8.26, observe that E is the mid-point ofAB, line l is passsing through E and is parallel to BCand CM || BA.

Prove that AF = CF by using the congruence of∆ AEF and ∆ CDF.

Example 7 : In ∆ ABC, D, E and F are respectivelythe mid-points of sides AB, BC and CA(see Fig. 8.27). Show that ∆ ABC is divided into fourcongruent triangles by joining D, E and F.Solution : As D and E are mid-points of sides ABand BC of the triangle ABC, by Theorem 8.9,

DE || ACSimilarly, DF || BC and EF || ABTherefore ADEF, BDFE and DFCE are all parallelograms.Now DE is a diagonal of the parallelogram BDFE,therefore, ∆ BDE ≅ ∆ FEDSimilarly ∆ DAF ≅ ∆ FEDand ∆ EFC ≅ ∆ FEDSo, all the four triangles are congruent.

Example 8 : l, m and n are three parallel linesintersected by transversals p and q such that l, mand n cut off equal intercepts AB and BC on p(see Fig. 8.28). Show that l, m and n cut off equalintercepts DE and EF on q also.Solution : We are given that AB = BC and haveto prove that DE = EF.Let us join A to F intersecting m at G..The trapezium ACFD is divided into two triangles;

Fig. 8.26

Fig. 8.27

Fig. 8.28

150 MATHEMATICS

namely ∆ ACF and ∆ AFD.In ∆ ACF, it is given that B is the mid-point of AC (AB = BC)and BG || CF (since m || n).So, G is the mid-point of AF (by using Theorem 8.10)Now, in ∆ AFD, we can apply the same argument as G is the mid-point of AF,GE || AD and so by Theorem 8.10, E is the mid-point of DF,i.e., DE = EF.In other words, l, m and n cut off equal intercepts on q also.

EXERCISE 8.21. ABCD is a quadrilateral in which P, Q, R and S are

mid-points of the sides AB, BC, CD and DA(see Fig 8.29). AC is a diagonal. Show that :

(i) SR || AC and SR = 12

AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

2. ABCD is a rhombus and P, Q, R and S are ©wthe mid-points of the sides AB, BC, CDand DA respectively. Show that the quadrilateral PQRS is a rectangle.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DArespectively. Show that the quadrilateral PQRS is a rhombus.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show thatF is the mid-point of BC.

Fig. 8.30

Fig. 8.29

QUADRILATERALS 151

5. In a parallelogram ABCD, E and F are themid-points of sides AB and CD respectively(see Fig. 8.31). Show that the line segments AFand EC trisect the diagonal BD.

6. Show that the line segments joining the mid-points of the opposite sides of aquadrilateral bisect each other.

7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse ABand parallel to BC intersects AC at D. Show that

(i) D is the mid-point of AC (ii) MD ⊥ AC

(iii) CM = MA = 12

AB

8.7 SummaryIn this chapter, you have studied the following points :1. Sum of the angles of a quadrilateral is 360°.2. A diagonal of a parallelogram divides it into two congruent triangles.3. In a parallelogram,

(i) opposite sides are equal (ii) opposite angles are equal(iii) diagonals bisect each other

4. A quadrilateral is a parallelogram, if(i) opposite sides are equal or (ii) opposite angles are equalor (iii) diagonals bisect each otheror (iv)a pair of opposite sides is equal and parallel

5. Diagonals of a rectangle bisect each other and are equal and vice-versa.6. Diagonals of a rhombus bisect each other at right angles and vice-versa.7. Diagonals of a square bisect each other at right angles and are equal, and vice-versa.8. The line-segment joining the mid-points of any two sides of a triangle is parallel to the

third side and is half of it.9. A line through the mid-point of a side of a triangle parallel to another side bisects the third

side.10. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral, in order,

is a parallelogram.

Fig. 8.31

EXERCISE 8.1Q.1. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the

angles of the quadrilateral.Sol. Suppose the measures of four angles are 3x, 5x, 9x and 13x.

∴ 3x + 5x + 9x + 13x = 360° [Angle sum property of a quadrilateral]⇒ 30x = 360°

⇒ x = 36030

° = 12°

⇒ 3x = 3 × 12° = 36°5x = 5 × 12° = 60°9x = 9 × 12° = 108°

13x = 13 × 12° = 156°∴ the angles of the quadrilateral are 36°, 60°, 108° and 156° Ans.

Q.2. If the diagonals of a parallelogram are equal, then show that it is arectangle.

Sol. Given : ABCD is a parallelogram in which AC = BD.To Prove : ABCD is a rectangle.Proof : In ∆ABC and ∆ABD

AB = AB [Common]

BC = AD[Opposite sides of a parallelogram]

AC = BD [Given]∴ ∆ABC ≅ ∆BAD [SSS congruence]

∠ABC = ∠BAD …(i) [CPCT]Since, ABCD is a parallelogram, thus,∠ABC + ∠BAD = 180° …(ii)

[Consecutive interior angles]∠ABC + ∠ABC = 180°

∴ 2∠ABC = 180° [From (i) and (ii)] ⇒ ∠ABC = ∠BAD = 90°This shows that ABCD is a parallelogram one of whose angle is 90°.Hence, ABCD is a rectangle. Proved.

Q.3. Show that if the diagonals of a quadrilateral bisect each other at rightangles, then it is a rhombus.

Sol. Given : A quadrilateral ABCD, in which diagonals AC and BD bisect eachother at right angles.To Prove : ABCD is a rhombus.

888 QUADRILATERALS

in

Proof : In ∆AOB and ∆BOCAO = OC[Diagonals AC and BD bisect each other]

∠AOB = ∠COB [Each = 90°]BO = BO [Common]

∴ ∆AOB ≅ ∆BOC [SAS congruence]AB = BC …(i) [CPCT]

Since, ABCD is a quadrilateral in whichAB = BC [From (i)]

Hence, ABCD is a rhombus.[∵ if the diagonals of a quadrilateral bisect each other, then it is a

parallelogram and opposite sides of a parallelogram are equal] Proved.

Q.4. Show that the diagonals of a square are equal and bisect each other atright angles.

Sol. Given : ABCD is a square in which AC and BD are diagonals.To Prove : AC = BD and AC bisects BD at right angles, i.e. AC ⊥ BD.AO = OC, OB = ODProof : In ∆ABC and ∆BAD,

AB = AB [Common]BC = AD [Sides of a square]

∠ABC = ∠BAD = 90° [Angles of a square]∴ ∆ABC ≅ ∆BAD [SAS congruence]⇒ AC = BD [CPCT]Now in ∆AOB and ∆COD,

AB = DC [Sides of a square]∠AOB = ∠COD [Vertically opposite angles]∠OAB = ∠OCD [Alternate angles]

∴ ∆AOB ≅ ∆COD [AAS congruence]∠AO = ∠OC [CPCT]

Similarly by taking ∆AOD and ∆BOC, we can show that OB = OD.In ∆ABC, ∠BAC + ∠BCA = 90° [ ∠B = 90°]⇒ 2∠BAC = 90° [∠BAC = ∠BCA, as BC = AD]⇒ ∠BCA = 45° or ∠BCO = 45°Similarly ∠CBO = 45°In ∆BCO.∠BCO + ∠CBO + ∠BOC = 180°⇒ 90° + ∠BOC = 180°⇒ ∠BOC = 90°⇒ BO ⊥ OC ⇒ BO ⊥ ACHence, AC = BD, AC ⊥ BD, AO = OC and OB = OD. Proved.

Q.5. Show that if the diagonals of a quadrilateral are equal and bisect eachother at right angles, then it is a square.

Sol. Given : A quadrilateral ABCD, in whichdiagonals AC and BD are equal and bisecteach other at right angles,To Prove : ABCD is a square.

in

Proof : Since ABCD is a quadrilateral whose diagonals bisect each other,so it is a parallelogram. Also, its diagonals bisect each other at rightangles, therefore, ABCD is a rhombus.⇒ AB = BC = CD = DA [Sides of a rhombus]In ∆ABC and ∆BAD, we have

AB = AB [Common]BC = AD [Sides of a rhombus]AC = BD [Given]

∴ ∆ABC ≅ ∆BAD [SSS congruence]∴ ∠ABC = ∠BAD [CPCT]

But, ∠ABC + ∠BAD = 180° [Consecutive interior angles]∠ABC = ∠BAD = 90°

∠A = ∠B = ∠C = ∠D = 90° [Opposite angles of a ||gm]⇒ ABCD is a rhombus whose angles are of 90° each.Hence, ABCD is a square. Proved.

Q.6. Diagonal AC of a parallelogram ABCDbisects ∠A (see Fig.). Show that

(i) it bisects ∠C also,(ii) ABCD is a rhombus.

Given : A parallelogram ABCD, in whichdiagonal AC bisects ∠A, i.e., ∠DAC = ∠BAC.

To Prove : (i) Diagonal AC bisects∠C i.e., ∠DCA = ∠BCA

(ii) ABCD is a rhomhus.Proof : (i) ∠DAC = ∠BCA [Alternate angles]

∠BAC = ∠DCA [Alternate angles]But, ∠DAC = ∠BAC [Given]∴ ∠BCA = ∠DCAHence, AC bisects ∠DCBOr, AC bisects ∠C Proved.

(ii) In ∆ABC and ∆CDAAC = AC [Common]

∠BAC = ∠DAC [Given]and ∠BCA = ∠DAC [Proved above]∴ ∆ABC ≅ ∆ADC [ASA congruence]∴ BC = DC [CPCT]But AB = DC [Given]∴ AB = BC = DC = ADHence, ABCD is a rhombus Proved.

[∵ opposite angles are equal]Q.7. ABCD is a rhombus. Show that diagonal AC bisects ∠ A as well as ∠C and

diagonal BD bisects ∠B as well as ∠D.Sol. Given : ABCD is a rhombus, i.e.,

AB = BC = CD = DA.To Prove : ∠DAC = ∠BAC,

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∠BCA = ∠DCA∠ADB = ∠CDB, ∠ABD = ∠CBDProof : In ∆ABC and ∆CDA, we have

AB = AD [Sides of a rhombus]AC = AC [Common]BC = CD [Sides of a rhombus]

∆ABC ≅ ∆ADC [SSS congruence]So, ∠DAC = ∠BAC

∠BCA = ∠DCASimilarly, ∠ADB = ∠CDB and ∠ABD = ∠CBD.Hence, diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as

well as ∠D. Proved.Q.8. ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠C.

Show that :(i) ABCD is a square (ii) diagonal BD bisects ∠B as well as ∠D.

Sol. Given : ABCD is a rectangle in which diagonal AC bisects ∠A as well as∠C.

To Prove : (i) ABCD is a square.(ii) Diagonal BD bisects ∠B as

well as ∠D.Proof : (i) In ∆ABC and ∆ADC, we have

∠BAC = ∠DAC [Given]∠BCA = ∠DCA [Given]

AC = AC∴ ∆ABC ≅ ∆ADC [ASA congruence]∴ AB = AD and CB = CD [CPCT]But, in a rectangle opposite sides are equal,i.e., AB = DC and BC = AD∴ AB = BC = CD = DAHence, ABCD is a square Proved.

(ii) In ∆ABD and ∆CDB, we haveAD = CDAB = CD [Sides of a square]BD = BD [Common]

∴ ∆ABD ≅ ∆CBD [SSS congruence]So, ∠ABD = ∠CBD

∠ADB = ∠CDBHence, diagonal BD bisects ∠B as well as ∠D Proved.

Q.9. In parallelogram ABCD, two points P and Q are taken on diagonal BDsuch that DP = BQ (see Fig.). Show that :

(i) ∆ APD ≅ ∆CQB(ii) AP = CQ

(iii) ∆ AQB ≅ ∆CPD(iv) AQ = CP(v) APCQ is a parallelogram

⎫⎬⎭

[CPCT]

⎫⎬⎭

⎫⎬⎭

[CPCT]

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Sol. Given : ABCD is a parallelogram and P and Q arepoints on diagonal BD such that DP = BQ.

To Prove : (i) ∆APD ≅ ∆CQB(ii) AP = CQ

(iii) ∆AQB ≅ ∆CPD(iv) AQ = CP(v) APCQ is a parallelogram

Proof : (i) In ∆APD and ∆CQB, we haveAD = BC [Opposite sides of a ||gm]DP = BQ [Given]

∠ADP = ∠CBQ [Alternate angles]∴ ∆APD ≅ ∆CQB [SAS congruence]

(ii) ∴ AP = CQ [CPCT](iii) In ∆AQB and ∆CPD, we have

AB = CD [Opposite sides of a ||gm]DP = BQ [Given]

∠ABQ = ∠CDP [Alternate angles]∴ ∆AQB ≅ ∆CPD [SAS congruence]

(iv) ∴ AQ = CP [CPCT](v) Since in APCQ, opposite sides are equal, therefore it

is a parallelogram. Proved.Q.10. ABCD is a parallelogram and AP and CQ

are perpendiculars from vertices A and Con diagonal BD (see Fig.). Show that

(i) ∆ APB ≅ ∆CQD(ii) AP = CQ

Sol. Given : ABCD is a parallelogram and APand CQ are perpendiculars from vertices Aand C on BD.

To Prove : (i) ∆APB ≅ ∆CQD(ii) AP = CQ

Proof : (i) In ∆APB and ∆CQD, we have∠ABP = ∠CDQ [Alternate angles]

AB = CD [Opposite sides of a parallelogram]∠APB = ∠CQD [Each = 90°]

∴ ∆APB ≅ ∆CQD [ASA congruence](ii) So, AP = CQ [CPCT] Proved.

Q.11. In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC= EF and BC || EF. Vertices A, B and C arejoined to vertices D, E and F respectively (seeFig.). Show that

(i) quadrilateral ABED is a parallelogram(ii) quadrilataeral BEFC is a parallelogram

(iii) AD || CF and AD = CF(iv) quadrilateral ACFD is a parallelogram(v) AC = DF

(vi) ∆ ABC ≡ ∆ DEF

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Sol. Given : In DABC and DDEF, AB = DE,AB ||DE, BC = EF and BC || EF. Vertices A, Band C are joined to vertices D, E and F.

To Prove : (i) ABED is a parallelogram(ii) BEFC is a parallelogram

(iii) AD || CF and AD = CF(iv) ACFD is a parallelogram(v) AC = DF

(vi) ∆ABC ≅ ∆DEFProof : (i) In quadrilateral ABED, we have

AB = DE and AB || DE. [Given]⇒ ABED is a parallelogram.

[One pair of opposite sides is parallel and equal](ii) In quadrilateral BEFC, we have

BC = EF and BC || EF [Given]⇒ BEFC is a parallelogram.

[One pair of opposite sides is parallel and equal](iii) BE = CF and BE||BECF [BEFC is parallelogram]

AD = BE and AD||BE [ABED is a parallelogram]⇒ AD = CF and AD||CF

(iv) ACFD is a parallelogram.[One pair of opposite sides is parallel and equal]

(v) AC = DF [Opposite sides of parallelogram ACFD](vi) In ∆ABC and ∆DEF, we have

AB = DE [Given]BC = EF [Given]AC = DF [Proved above]

∴ ∆ABC ≅ ∆DEF [SSS axiom] Proved.Q.12. ABCD is a trapezium in which AB

|| CD and AD = BC (see Fig.).Show that

(i) ∠ A = ∠ B(ii) ∠C = ∠D

(iii) ∆ ABC ≅ ∆ BAD(iv) diagonal AC = diagonal BD

Sol. Given : In trapezium ABCD, AB || CD and AD = BC.To Prove : (i) ∠A = ∠B

(ii) ∠C = ∠D(iii) ∆ABC ≅ ∆BAD(iv) diagonal AC = diagonal BD

Constructions : Join AC and BD. Extend AB and drawa line through C parallel to DA meeting AB producedat E.

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Proof : (i) Since AB || DC⇒ AE || DC …(i)and AD || CE …(ii) [Construction]⇒ ADCE is a parallelogram [Opposite pairs of

sides are parallel∠A + ∠E = 180° …(iii)

[Consecutive interior angles]∠B + ∠CBE = 180° …(iv) [Linear pair]

AD = CE …(v) [Opposite sides of a ||gm]AD = BC …(vi) [Given]

⇒ BC = CE [From (v) and (vi)]⇒ ∠E = ∠CBE …(vii) [Angles opposite to

equal sides]∴ ∠B + ∠E = 180° …(viii) [From (iv) and (vii)Now from (iii) and (viii) we have

∠A + ∠E = ∠B + ∠E⇒ ∠A = ∠B Proved.

(ii) ∠A + ∠D = 180°∠B + ∠C = 180°

⇒ ∠A + ∠D = ∠B + ∠C [∵ ∠A = ∠B]⇒ ∠D = ∠COr ∠C = ∠D Proved.

(iii) In ∆ABC and ∆BAD, we haveAD = BC [Given]∠A = ∠B [Proved]AB = CD [Common]

∴ ∆ABC ≅ ∆BAD [ASA congruence](iv) diagonal AC = diagonal BD [CPCT] Proved.

⎫⎬⎭ [Consecutive interior angles]

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888 QUADRILATERALS

∴ PQ || AC and PQ = 12 AC …(1)

[Mid-point theorem]In ∆ADC, R is the mid-point of CD and S is the mid-pointof AD

∴ SR || AC and SR = 12 AC …(2)

[Mid-point theorem](ii) From (1) and (2), we get

PQ || SR and PQ = SR(iii) Now in quadrilateral PQRS, its one pair of opposite sides

PQ and SR is equal and parallel.∴ PQRS is a parallelogram. Proved.

EXERCISE 8.2Q.1. ABCD is a quadrilateral in which P, Q, R and S

are mid-points of the sides AB, BC, CD and DArespectively. (see Fig.). AC is a diagonal. Showthat :

(i) SR || AC and SR = 12 AC

(ii) PQ = SR(iii) PQRS is a parallelogram.

Given : ABCD is a quadrilateral in which P, Q, R and Sare mid-points of AB, BC, CD and DA. AC is a diagonal.

To Prove : (i) SR || AC and SR = 12 AC

(ii) PQ = SR(iii) PQRS is a parallelogram

Proof : (i) In ∆ABC, P is the mid-point of AB and Q is the mid-pointof BC.

Q.2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB,BC, CD and DA respectively. Show that the quadrilateral PQRS is arectangle.

Sol. Given : ABCD is a rhombus in which P, Q, R and S are mid points ofsides AB, BC, CD and DA respectively :To Prove : PQRS is a rectangle.Construction : Join AC, PR and SQ.Proof : In ∆ABCP is mid point of AB [Given]Q is mid point of BC [Given]

⇒ PQ || AC and PQ = 12 AC …(i) [Mid point theorem]

Similarly, in ∆DAC,

SR || AC and SR = 12 AC …(ii)

From (i) and (ii), we have PQ||SR and PQ = SR⇒ PQRS is a parallelogram

[One pair of opposite sides is parallel and equal]Since ABQS is a parallelogram⇒ AB = SQ [Opposite sides of a || gm]Similarly, since PBCR is a parallelogram.⇒ BC = PRThus, SQ = PR [AB = BC]Since SQ and PR are diagonals of parallelogram PQRS, which are equal.⇒ PQRS is a rectangle. Proved.

Q.3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC,CD and DA respectively. Show that the quadrilataral PQRS is a rhombus.

Sol. Given : A rectangle ABCD in which P, Q, R, S arethe mid-points of AB, BC, CD and DA respectively,PQ, QR, RS and SP are joined.To Prove : PQRS is a rhombus.Construction : Join AC

Proof : In ∆ABC, P and Q are the mid-points of the sides AB and BC.

∴ PQ || AC and PQ = 12 AC …(i) [Mid point theorem]

Similarly, in ∆ADC,

SR || AC and SR = 12 AC …(ii)

From (i) and (ii), we getPQ || SR and PQ = SR …(iii)Now in quadrilateral PQRS, its one pair of opposite sides PQ and SR isparallel and equal [From (iii)]∴PQRS is a parallelogram.

Now AD = BC …(iv)[Opposite sides of a rectangle ABCD]

∴ 12 AD = 1

2 BC

⇒ AS = BQ In ∆APS and ∆BPQ

AP = BP [∵ P is the mid-point of AB]AS = BQ [Proved above]

∠PAS = ∠PBQ [Each = 90°]∆APS ≅ ∆BPQ [SAS axiom]

∴ PS = PQ …(v) From (iii) and (v), we have PQRS is a rhombus Proved.

Q.4. ABCD is a trapezium in whichAB || DC, BD is a diagonal and E is themid-point of AD. A line is drawn through Eparallel to AB intersecting BC at F (seeFig.). Show that F is the mid-point of BC.

Sol. Given : A trapezium ABCD withAB || DC, E is the mid-point of AD and EF|| AB.To Prove : F is the mid-point of BC.Proof : AB || DC and EF || AB⇒ AB, EF and DC are parallel.Intercepts made by parallel lines AB, EF and DC on transversal AD areequal.∴ Intercepts made by those parallel lines on transversal BC are alsoequal.i.e., BF = FC⇒ F is the mid-point of BC.

Q.5. In a parallelogram ABCD, E and F are themid-points of sides AB and CD respectively(see Fig.). Show that the line segments AFand EC trisect the diagonal BD.

O

Sol. Given : A parallelogram ABCD, in whichE and F are mid-points of sides AB and DCrespectively.To Prove : DP = PQ = QBProof : Since E and F are mid-points of AB and DC respectively.

⇒ AE = 12 AB and CF = 1

2 DC …(i)

But, AB = DC and AB || DC …(ii)[Opposite sides of a parallelogram]

∴ AE = CF and AE || CF.⇒ AECF is a parallelogram.

[One pair of opposite sides is parallel and equal]In ∆BAP,E is the mid-point of ABEQ || AP⇒ Q is mid-point of PB [Converse of mid-point theorem]⇒ PQ = QB …(iii)Similarly, in ∆DQC,P is the mid-point of DQ

DP = PQ …(iv)From (iii) and (iv), we haveDP = PQ = QBor line segments AF and EC trisect the diagonal BD. Proved.

Q.6. Show that the line segments joining the mid-points of the opposite sides ofa quadrilateral bisect each other.

Sol. Given : ABCD is a quadrilateral in which EG andFH are the line segments joining the mid-points ofopposite sides.To Prove : EG and FH bisect each other.Construction : Join EF, FG, GH, HE and AC.Proof : In ∆ABC, E and F are mid-points of AB and BC respectively.

∴ EF = 12 AC and EF || AC …(i)

In ∆ADC, H and G are mid-points of AD and CD respectively.

∴ HG = 12 AC and HG || AC …(ii)

From (i) and (ii), we getEF = HG and EF || HG∴ EFGH is a parallelogram.

[∵ a quadrilateral is a parallelogram if itsone pair of opposite sides is equal and parallel]

Now, EG and FH are diagonals of the parallelogram EFGH.∴ EG and FH bisect each other.

[Diagonal of a parallelogram bisect each other] Proved.

Q.7. ABC is a triangle right angled at C. A line through the mid-point M ofhypotenuse AB and parallel to BC intersects AC at D. Show that

(i) D is the mid-point of AC.(ii) MD ⊥ AC

(iii) CM = MA = 12 AB

Sol. Given : A triangle ABC, in which ∠C = 90° and M is the mid-point of ABand BC || DM.

To Prove : (i) D is the mid-point of AC[Given]

(ii) DM ⊥ BC

(iii) CM = MA = 12 AB

Construction : Join CM.Proof : (i) In ∆ABC,

M is the mid-point of AB. [Given]BC || DM [Given]D is the mid-point of AC

[Converse of mid-point theorem] Proved.(ii) ∠ADM = ∠ACB [∵ Coresponding angles]

But ∠ACB = 90° [Given]∴ ∠ADM = 90°But ∠ADM + ∠CDM = 180° [Linear pair]∴ ∠CDM = 90°Hence, MD ⊥ AC Proved.

(iii) AD = DC …(1) [∵ D is the mid-point of AC]Now, in ∆ADM and ∆CMD, we have

∠ADM = ∠CDM [Each = 90°]AD = DC [From (1)]DM = DM [Common]

∴ ∆ADM ≅ ∆CMD [SAS congruence]⇒ CM = MA …(2) [CPCT]Since M is mid-point of AB,

∴ MA = 12 AB …(3)

Hence, CM = MA = 12 AB Proved. [From (2) and (3)]