# Important Questions for CBSE Class 9 Maths Chapter 8 - Quadrilaterals

## CBSE Class 9 Maths Chapter-8 Important Questions - Free PDF Download

**1 Marks Quetions**

**1. A quadrilateral ABCD is a parallelogram if **

**(a) AB = CD **

**(b) ABBC**

**(c) **

**(d) AB = AD**

**Ans. (c) **

**2. In figure, ABCD and AEFG are both parallelogram if **

**(a) **

**(b) **

**(c) **

**(d) **

**Ans. (c) **

**3. In a square ABCD, the diagonals AC and BD bisects at O. Then is **

**(a) acute angled **

**(b) obtuse angled**

**(c) equilateral **

**(d) right angled **

**Ans. (d) **right angled

**4. ABCD is a rhombus. If **

**(a) **

**(b) **

**(c) **

**(d) **

**Ans. (c) **

**5. In fig ABCD is a parallelogram. If and then is **

**Ans. **

**6. If the diagonals of a quadrilateral bisect each other, then the quadrilateral must be. **

**(a) Square **

**(b) Parallelogram **

**(c) Rhombus **

**(d) Rectangle **

**Ans. (b) **Parallelogram

**7. The diagonal AC and BD of quadrilateral ABCD are equal and are perpendicular bisector of each other then quadrilateral ABCD is a **

**(a) Kite **

**(b) Square **

**(c) Trapezium **

**(d) Rectangle **

**Ans. (b) **Square

**8. The quadrilateral formed by joining the mid points of the sides of a quadrilateral ABCD taken in order, is a rectangle if **

**(a) ABCD is a parallelogram **

**(b) ABCD is a rut angle **

**(c) Diagonals AC and BD are perpendicular **

**(d) AC=BD **

**Ans. (a) **ABCD is a parallelogram

**9. In the fig ABCD is a Parallelogram. The values of and are**

**(a) 30, 35 **

**(b) 45, 30**

**(c) 45, 45 **

**(d) 55, 35**

**Ans. (b) **45, 30

**10. In fig if DE=8 cm and D is the mid-Point of AB, then the true statement is **

** **

**(a) AB=AC **

**(b) DE||BC **

**(c) E is not mid-Point of AC **

**(d) DEBC**

**Ans. (c) **E is not mid-Point of AC

**11. The sides of a quadrilateral extended in order to form exterior angler. The sum of these exterior angle is **

**(a) **

**(b) **

**(c) **

**(d) **

**Ans. (d) **

**12. ABCD is rhombus with The measure of is **

**(a) **

**(b) **

**(c) **

**(d) **

**Ans. b) **

**13. In fig D is mid-point of AB and DEBC then AE is equal to **

**(a) AD **

**(b) EC**

**(c) DB **

**(d) BC **

**Ans. (b) **EC

**14. In fig D and E are mid-points of AB and AC respectively. The length of DE is**

**(a) 8.2 cm **

**(b) 5.1 cm **

**(c) 4.9 cm **

**(d) 4.1 cm**

**Ans. (d) **4.1 cm

**15. A diagonal of a parallelogram divides it into **

**(a) two congruent triangles **

**(b) two similes triangles **

**(c) two equilateral triangles **

**(d) none of these **

**Ans. (a) **two congruent triangles

**16. A quadrilateral is a _________, if its opposite sides are equal:**

**(a) Kite **

**(b) trapezium **

**(c) cyclic quadrilateral **

**(d) parallelogram **

**Ans. (d) **parallelogram

**17. In the adjoining Fig. AB = AC. CD||BA and AD is the bisector of prove that **

**(a) and**

**Ans. **In

[Opposite angle of equal sides are equal]

[Exterior angle]

Now

Also CD||BA Given)

is a parallelogram

(ii) ABCD is a parallelogram

**18. Which of the following is not a parallelogram?**

**(a) Rhombus **

**(b) Square **

**(c) Trapezium **

**(d) Rectangle**

**Ans. (c)** Trapezium

**19. The sum of all the four angles of a quadrilateral is **

**(a) 180 ^{0 }**

**(b) 360 ^{0 } **

**(c) 270 ^{0} **

**(d) 90 ^{0}**

**Ans. (b) **360^{0}

**20. In Fig ABCD is a rectangle P and Q are mid-points of AD and DC respectively. Then length of PQ is **

**(a)5 cm **

**(b) 4 cm **

**(c) 2.5 cm **

**(d) 2 cm**

**Ans. (c) **2.5 cm

**21. In Fig ABCD is a rhombus. Diagonals AC and BD intersect at O. E and F are mid points of AO and BO respectively. If AC = 16 cm and BD = 12 cm then EF is **

**(a)10 cm **

**(b) 5 cm **

**(c) 8 cm **

**(d) 6 cm**

**Ans. (b) **5 cm

**2 Marks Quetions**

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

**Ans. **Let in quadrilateral ABCD, A = B = C = and D =

Since, sum of all the angles of a quadrilateral =

A + B + C + D =

Now A =

B =

C =

And D =

Hence angles of given quadrilateral are and

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

**Ans. **Given: ABCD is a parallelogram with diagonal AC = diagonal BD

To prove: ABCD is a rectangle.

Proof: In triangles ABC and ABD,

AB = AB [Common]

AC = BD [Given]

AD = BC [opp. Sides of a gm]

ABC BAD [By SSS congruency]

DAB = CBA [By C.P.C.T.] ……….(i)

But DAB + CBA = ……….(ii)

[ ADBC and AB cuts them, the sum of the interior angles of the same side of transversal is ]

From eq. (i) and (ii),

DAB = CBA =

Hence ABCD is a rectangle.

**3. Diagonal AC of a parallelogram ABCD bisects ****A (See figure). Show that:**

**(i) It bisects ****C also.**

**(ii) ABCD is a rhombus.**

**Ans. **Diagonal AC bisects A of the parallelogram ABCD.

**(i) **Since AB DC and AC intersects them.

1 = 3 [Alternate angles] ……….(i)

Similarly 2 = 4 ……….(ii)

But 1 = 2 [Given] ……….(iii)

3 = 4 [Using eq. (i), (ii) and (iii)]

Thus AC bisects C.

**(ii) **2 = 3 = 4 = 1

AD = CD [Sides opposite to equal angles]

AB = CD = AD = BC

Hence ABCD is a rhombus.

**4. ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices A and C on its diagonal BD (See figure). Show that:**

**(i) ****APB ****CQD**

**(ii) AP = CQ**

**Ans**. Given: ABCD is a parallelogram. AP BD and CQ BD

To prove: (i) APB CQD (ii) AP = CQ

Proof: (i) InAPB and CQD,

1 = 2 [Alternate interior angles]

AB = CD [Opposite sides of a parallelogram are equal]

APB = CQD =

APB CQD [By ASA Congruency]

(ii) Since APB CQD

AP = CQ [By C.P.C.T.]

**5. ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively (See figure). AC is a diagonal. Show that:**

**(i) SR **** AC and SR = **** AC**

**(ii) PQ = SR**

**(iii) PQRS is a parallelogram.**

**Ans**. In ABC, P is the mid-point of AB and Q is the mid-point of BC.

Then PQ AC and PQ = AC

**(i) **In ACD, R is the mid-point of CD and S is the mid-point of AD.

Then SR AC and SR = AC

**(ii) **Since PQ = AC and SR = AC

Therefore, PQ = SR

**(iii) **Since PQ AC and SR AC

Therefore, PQ SR [two lines parallel to given line are parallel to each other]

Now PQ = SR and PQ SR

Therefore, PQRS is a parallelogram.

**6. The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.**

**Ans. **Suppose angles of quadrilateral ABCD are 3x, 5x, 9x, and 13x

[sum of angles of a quadrilateral is ]

**7. Show that each angle of a rectangle is a right angle.**

**Ans. **We know that rectangle is a parallelogram whose one angle is right angle.

Let ABCD be a rectangle.

To prove

Proof: and AB is transversal

**8. A transversal cuts two parallel lines prove that the bisectors of the interior angles enclose a rectangle.**

**Ans. ** and EF cuts them at P and R.

[alternate interior angles]

[alternate]

**9. Prove that diagonals of a rectangle are equal in length.**

**Ans. **ABCD is a rectangle and AC and BD are diagonals.

To prove AC = BD

Proof:

AD = BC [In a rectangle opposite sides are equal]

AB = AB common [common]

**10. If each pair of opposite sides of a quadrilateral is equal, then prove that it is a parallelogram. **

**Ans. **Given A quadrilateral ABCD in which AB = DC and AD = BC

To prove: ABCD is a parallelogram

Construction: Join AC

__Proof:__ In and

AD=BC (Given)

AB=DC

AC=AC [common]

is a parallelogram.

**11. **

**Ans. **ABCD is a parallelogram. The diagonals of a parallelogram bisect bisect each other

But [given]

Or OX=OY

Now in quadrilateral AYCX, the diagonals AC and XY bisect each other

is a parallelogram.

In fig ABCD is a parallelogram and x, y are the points on the diagonal BD such that Dx<By show that AYCX is a parallelogram.

**12. Show that the line segments joining the mid points of opposite sides of a quadrilateral bisect each other. **

**Ans. **Given ABCD is quadrilateral E, F, G, H are mid points of the side AB, BC, CD and DA respectively

To prove: EG and HF bisect each other.

In , E is mid-point of AB and F is mid-point of BC

And

Similarly, and

From (i) and (ii), and

is a parallelogram and EG and HF are its diagonals

Diagonals of a parallelogram bisect each other

Thus, EG and HF bisect each other.

**13. ABCD is a rhombus show that diagonal AC bisects as well as and diagonal BD bisects as well as **

**Ans. **ABCD is a rhombus

In and

[Sides of a rhombus]

[Sides of a rhombus]

[Common]

[By SSS Congruency]

And

Hence AC bisects as well as

Similarly, by joining B to D, we can prove that

Hence BD bisects as well as

**14. In fig AD is a median of is mid-Point of AD.BE produced meet AC at F. Show that **

**Ans. **Let M is mid-Point of CF Join DM

In is mid- Point of AD and

is mid-point of AM

FM=MC

Hence Proved.

**15. Prove that a quadrilateral is a parallelogram if the diagonals bisect each other.**

**Ans. **ABCD is a quadrilateral in which diagonals AC and BD intersect each other at O

In and

[Given]

[Given]

And [Vertically apposite angle

[By SAS]

[By C.P.C.T]

But this is Pair of alternate interior angles

Similarly AD||BC

Quadrilateral ABCD is a Parallelogram.

**16. In fig ABCD is a Parallelogram. AP and CQ are Perpendiculars from the Vertices A and C on diagonal BD.**

**Show that **

**(i) **

**(ii) **

**Ans. (I) **in and

AB=DC [opposite sides of a Parallelogram]

And

[ASA]

**(II) **(By C.P.C.T)

**17. ABCD is a Parallelogram E and F are the mid-Points of BC and AD respectively. Show that the segments BF and DE trisect the diagonal AC.**

**Ans. **FD||BE and FD=BE

BEDF Is a Parallelogram

EG||BH and E is the mid-Point of BC

G is the mid-point of HC

Or HG=GC…………..(i)

Similarly AH=HG………….(ii)

From (i) and (ii) we get

AH=HG=GC

Thus the segments BF and DE bisects the diagonal AC.

**18. Prove that if each pair of apposite angles of a quadrilateral is equal, then it is a parallelogram. **

**Ans. **Given: ABCD is a quadrilateral in which and

To Prove: ABCD is a parallelogram

Proof: [Given]

[Given]

In quadrilateral. ABCD

[By….(i)]

These are sum of interior angles on the same side of transversal

and

ABCD is a parallelogram**.**

**19. In Fig. ABCD is a trapezium in which AB||DC E is the mid-point of AD. A line through E is parallel to AB show that bisects the side BC**

**Ans. **Join AC

In

E is mid-point of AD and EO||DC

O is mid point of AC [A line segment joining the midpoint of one side of a parallel to second side and bisect the third side]

In

O is mid point of AC

OF||AB F is mid point of BC

Bisect BC

**20. In Fig. ABCD is a parallelogram in which X and Y are the mid-points of the sides DC and AB respectively. Prove that AXCY is a parallelogram**

**Ans. **In the given fig

ABCD is a parallelogram

AB||CD and AB = CD

And

And

[X and Y are mid-point of DC and AB respectively]

is a parallelogram

**21. The angles of quadrilateral are in the ratio 3:5:10:12 Find all the angles of the quadrilateral. **

**Ans. **Suppose angles of quadrilaterals are

In a quadrilateral

30x=360

**22. In fig D is mid-points of AB. P is on AC such that and DE||BP show that **

**Ans. **In ABP

D is mid points of AB and DE||BP

E is midpoint of AP

AE = EP also PC = AP

2PC = AP

2PC = 2AE

PC = AE

AE = PE = PC

AC = AE + EP + PC

AC = AE + AE + AE

AE =AC

Hence Proved.

**23. Prove that the bisectors of the angles of a Parallelogram enclose a rectangle. It is given that adjacent sides of the parallelogram are unequal.**

**Ans. **ABCD is a parallelogram

Or [Sum of angle of a ]

Similarly, and

. Thus each angle of quadrilateral PQRS is

Hence PQRS is a rectangle.

**24. Prove that a quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal**

**Ans**. Given: ABCD is a quadrilateral in which AB||DC and BC||AD.

To Prove: ABCD is a parallelogram

Construction: Join AC and BD intersect each other at O.

Proof: [By AAA

Because

AO=OC

And BO=OD

ABCD is a parallelogram

Diagonals of a parallelogram bisect each other.

**3 Marks Quetions**

**1. Show that is diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.**

**Ans. **Given: Let ABCD is a quadrilateral.

Let its diagonal AC and BD bisect each other at right angle at point O.

OA = OC, OB = OD

AndAOB = BOC = COD = AOD =

To prove: ABCD is a rhombus.

Proof: In AOD and BOC,

OA = OC[Given]

AOD = BOC[Given]

OB = OD[Given]

AODCOB [By SAS congruency]

AD = CB [By C.P.C.T.]……….(i)

Again, In AOB and COD,

OA = OC[Given]

AOB = COD[Given]

OB = OD[Given]

AOBCOD [By SAS congruency]

AD = CB[By C.P.C.T.]……….(ii)

Now In AOD and BOC,

OA = OC[Given]

AOB = BOC[Given]

OB = OB[Common]

AOBCOB [By SAS congruency]

AB = BC [By C.P.C.T.]……….(iii)

From eq. (i), (ii) and (iii),

AD = BC = CD = AB

And the diagonals of quadrilateral ABCD bisect each other at right angle.

Therefore, ABCD is a rhombus.

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

**Ans. **Given: ABCD is a square. AC and BD are its diagonals bisect each other at point O.

To prove: AC = BD and AC BD at point O.

Proof: In triangles ABC and BAD,

AB = AB[Common]

ABC = BAD =

BC = AD [Sides of a square]

ABC BAD [By SAS congruency]

AC = BD [By C.P.C.T.] Hence proved.

Now in triangles AOB and AOD,

AO = AO[Common]

AB = AD[Sides of a square]

OB = OD[Diagonals of a square bisect each other]

AOBAOD[By SSS congruency]

AOB = AOD[By C.P.C.T.]

ButAOB + AOD = [Linear pair]

AOB = AOD =

OA BD or AC BD

Hence proved.

**3. ABCD is a rhombus. Show that the diagonal AC bisects A as well as C and diagonal BD bisects B as well as D.**

**Ans**. ABCD is a rhombus. Therefore, AB = BC = CD = AD

Let O be the point of bisection of diagonals.

OA = OC and OB = OD

In AOB and AOD,

OA = OA[Common]

AB = AD[Equal sides of rhombus]

OB = OD(diagonals of rhombus bisect each other]

AOBAOD[By SSS congruency]

OAD = OAB[By C.P.C.T.]

OA bisects A……….(i)

SimilarlyBOC DOC[By SSS congruency]

OCB = OCD[By C.P.C.T.]

OC bisects C……….(ii)

From eq. (i) and (ii), we can say that diagonal AC bisects A and C.

Now in AOB and BOC,

OB = OB[Common]

AB = BC[Equal sides of rhombus]

OA = OC(diagonals of rhombus bisect each other]

AOBCOB[By SSS congruency]

OBA = OBC[By C.P.C.T.]

OB bisects B……….(iii)

SimilarlyAODCOD[By SSS congruency]

ODA = ODC[By C.P.C.T.]

BD bisects D……….(iv)

From eq. (iii) and (iv), we can say that diagonal BD bisects B and D

**4. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (See figure). Show that:**

**(i) ****APD****CQB**

**(ii) AP = CQ**

**(iii) ****AQB****CPD**

**(iv) AQ = CP**

**(v) APCQ is a parallelogram. **

**Ans. (i)**In APD and CQB,

DP = BQ[Given]

ADP = QBC[Alternate angles (ADBC and BD is transversal)]

AD = CB[Opposite sides of parallelogram]

APDCQB[By SAS congruency]

**(ii) **SinceAPDCQB

AP = CQ[By C.P.C.T.]

**(iii) **In AQB and CPD,

BQ = DP[Given]

ABQ = PDC[Alternate angles (ABCD and BD is transversal)]

AB = CD[Opposite sides of parallelogram]

AQBCPD[By SAS congruency]

**(iv) **SinceAQBCPD

AQ = CP[By C.P.C.T.]

**(v)** In quadrilateral APCQ,

AP = CQ[proved in part (i)]

AQ = CP[proved in part (iv)]

Since opposite sides of quadrilateral APCQ are equal.

Hence APCQ is a parallelogram.

**5. ABCD is a rhombus and P, Q, R, S are mid-points of AB, BC, CD and DA respectively. Prove that quadrilateral PQRS is a rectangle.**

**Ans. **Given: P, Q, R and S are the mid-points of respective sides AB, BC, CD and DA of rhombus. PQ, QR, RS and SP are joined.

To prove: PQRS is a rectangle.

Construction: Join A and C.

Proof: In ABC, P is the mid-point of AB and Q is the mid-point of BC.

PQ AC and PQ = AC ……….(i)

In ADC, R is the mid-point of CD and S is the mid-point of AD.

SR AC and SR = AC……….(ii)

From eq. (i) and (ii),PQ SR and PQ = SR

PQRS is a parallelogram.

Now ABCD is a rhombus. [Given]

AB = BC

AB = BCPB = BQ

1 = 2[Angles opposite to equal sides are equal]

Now in triangles APS and CQR, we have,

AP = CQ[P and Q are the mid-points of AB and BC and AB = BC]

Similarly AS = CR and PS = QR[Opposite sides of a parallelogram]

APS CQR[By SSS congreuancy]

3 = 4[By C.P.C.T.]

Now we have1 + SPQ + 3 =

And2 + PQR + 4 = [Linear pairs]

1 + SPQ + 3 = 2 + PQR + 4

Since1 = 2 and 3 = 4[Proved above]

SPQ = PQR……….(iii)

Now PQRS is a parallelogram [Proved above]

SPQ + PQR = ……….(iv)[Interior angles]

Using eq. (iii) and (iv),

SPQ + SPQ = 2SPQ =

SPQ =

Hence PQRS is a rectangle.

**6. ABCD is a rectangle 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 a rhombus.**

**Ans. **Given: A rectangle ABCD in which P, Q, R and S are the mid-points of the sides 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 sides AB, BC respectively.

PQ AC and PQ = AC……….(i)

In ADC, R and S are the mid-points of sides CD, AD respectively.

SR AC and SR = AC……….(ii)

From eq. (i) and (ii), PQ SR and PQ = SR……….(iii)

PQRS is a parallelogram.

Now ABCD is a rectangle.[Given]

AD = BC

AD = BCAS = BQ……….(iv)

In triangles APS and BPQ,

AP = BP[P is the mid-point of AB]

PAS = PBQ[Each ]

And AS = BQ[From eq. (iv)]

APS BPQ[By SAS congruency]

PS = PQ[By C.P.C.T.]………(v)

From eq. (iii) and (v), we get that PQRS is a parallelogram.

PS = PQ

Two adjacent sides are equal.

Hence, PQRS is a rhombus.

**7. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (See figure). Show that the line segments AF and EC trisect the diagonal BD.**

**Ans. **Since E and F are the mid-points of AB and CD respectively.

AE = AB and CF = CD……….(i)

But ABCD is a parallelogram.

AB = CD and AB DC

AB = CD and AB DC

AE = FC and AE FC[From eq. (i)]

AECF is a parallelogram.

FA CEFP CQ[FP is a part of FA and CQ is a part of CE] ……...(ii)

Since the segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side.

In DCQ, F is the mid-point of CD and FP CQ

P is the mid-point of DQ.

DP = PQ……….(iii)

Similarly, In ABP, E is the mid-point of AB and EQ AP

Q is the mid-point of BP.

BQ = PQ……….(iv)

From eq. (iii) and (iv),

DP = PQ = BQ………(v)

Now BD = BQ + PQ + DP = BQ + BQ + BQ = 3BQ

BQ = BD……….(vi)

From eq. (v) and (vi),

DP = PQ = BQ = BD

Points P and Q trisects BD.

So AF and CE trisects BD.

**8. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.**

**Ans. (i) **In ABC, M is the mid-point of AB[Given]

MD BC

AD = DC[Converse of mid-point theorem]

Thus D is the mid-point of AC.** **

**(ii) ** BC (given) consider AC as a transversal.

1 = C[Corresponding angles]

1 = [C = ]

Thus MD AC.

**(iii) **In AMD and CMD,

AD = DC[proved above]

1 = 2 = [proved above]

MD = MD[common]

AMD CMD[By SAS congruency]

AM = CM[By C.P.C.T.]……….(i)

Given that M is the mid-point of AB.

AM = AB……….(ii)

From eq. (i) and (ii),

CM = AM = AB

**9. In a parallelogram ABCD, bisectors of adjacent angles A and B intersect each other at P. prove that **

**Ans. **Given ABCD is a parallelogram is and bisectors of intersect each other at P.** **

To prove

Proof:

But ABCD is a parallelogram and ADBC

Hence Proved

**10. In figure diagonal AC of parallelogram ABCD bisects show that **

**(i) if bisects **

**ABCD is a rhombus**

**Ans.(i) **ABDC and AC is transversal

(Alternate angles)

And (Alternate angles)

But,

** **

**(ii) **

AC=AC [common]

[given]

[proved]

[By CPCT]

**11. In figure ABCD is a parallelogram. AX and CY bisects angles A and C. prove that AYCX is a parallelogram. **

**Ans. **Given in a parallelogram AX and CY bisects respectively and we have to show that AYCX in a parallelogram.

…(i) [opposite angles of parallelogram]

[Given] …(ii)

And [give] …..(iii)

But

By (2) and (3), we get

Also, [opposite sides of parallelogram] ….(v)

From (i), (iv) and (v), we get

But, AB =CD [opposite sides of parallelogram]

AB-BY=CD-DX

Or

Ay=CX

But

is a parallelogram

**12. Prove that the line segment joining the mid-points of two sides of a triangle is parallel to the third side.**

**Ans. **Given ABC in which E and F are mid points of side AB and AC respectively.

To prove: EF||BC

Construction: Produce EF to D such that EF = FD. Join CD

Proof:

AF=FC

[vertically opposite angles]

EF=FD [By construction]

And

AE= BE[is the mid-point]

And

**13. Prove that a quadrilateral is a rhombus if its diagonals bisect each other at right angles.**

**Ans. **Given ABCD is a quadrilateral diagonals AC and BD bisect each other at O at right angles

__To Prove__: ABCD is a rhombus

__Proof:__ diagonals AC and BD bisect each other at O

And

Now In And

Given

[Common]

And (Given)

(SAS)

(C.P.C.T.)

Similarly, BC=CD, CD=DA and DA=AB,

Hence, ABCD is a rhombus.

**14. Prove that the straight line joining the mid points of the diagonals of a trapezium is parallel to the parallel sides.**

**Ans. **Given a trapezium ABCD in which and M,N are the mid Points of the diagonals AC and BD.

We need to prove that

Join CN and let it meet AB at E

Now in and

[Alternate angles]

[Alternate angles]

And [given]

[ASA]

[By C.P.C.T]

Now in and are the mid points of the sides AC and CE respectively.

Or

Also

**15. In fig is a right angle in is the mid-point of intersects BC at E. show that**

**(i) E is the mid-point of BC **

**(ii) DEBC **

**(ii) BD = AD**

**Ans. **__Proof:__and D is mid points of AC

In and

CE=BE

DE= DE

And

**16. ABC is a triangle and through vertices A, B and C lines are drawn parallel to BC, AC and AB respectively intersecting at D, E and F. prove that perimeter of is double the perimeter of .**

**Ans. **Is a parallelogram

Is a parallelogram

Or

Similarly, ED = 2AB and FD = 2AC

Perimeter of

Perimeter of

= 2AB+2BC+2AC

= 2[AB+BC+AC]

= 2 Perimeter of

Hence Proved.

**17. In fig ABCD is a quadrilateral P, Q, R and S are the mid Points of the sides AB, BC, CD and DA, AC is diagonal. Show that**

**(i) SR||AC **

**(ii) PQ=SR **

**(iii) PQRS is a parallelogram **

**(iv) PR and SQ bisect each other **

**Ans. **In ABC, P and Q are the mid-points of the sides AB and BC respectively

**(i)** PQ||AC and PQ=AC

**(ii)** Similarly SR||AC and SR=AC

PQ||SR and PQ=SR

**(iii)** Hence PQRS is a Parallelogram.

**(iv)** PR and SQ bisect each other.

**18. In are respectively the mid-Points of sides AB,DC and CA. show that is divided into four congruent triangles by Joining D,E,F.**

**Ans. **D and E are mid-Points of sides AB and BC of ABC

DE||AC {A line segment joining the mid-Point of any two sides of a triangle parallel to third side}

Similarly, DF||BC and EF||AB

ADEF, BDEF and DFCE are all Parallelograms.

DE is diagonal of Parallelogram BDFE

Similarly, DAFFED

And EFCFED

So all triangles are congruent

**19. ABCD is a Parallelogram is which P and Q are mid-points of opposite sides AB and CD. If AQ intersect DP at S BQ intersects CP at R, show that**

**(i) APCQ is a Parallelogram**

**(ii) DPBQ is a parallelogram **

**(iv) PSQR is a parallelogram**

**Ans**. **(i)** In quadrilateral APCQ

AP||QC [AB||CD]……..(i)

AP=AB, CQ=CD (Given)

Also AB= CD

So AP=QC……….(ii)

Therefore, APCQ is a parallelogram

[It any two sides of a quadrilateral equal and parallel then quad is a parallelogram]

**(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||PR

So. PSQR is also parallelogram.

**20.are three parallel lines intersected by transversals P and q such that and cut off equal intercepts AB and BC on P In fig Show that cut off equal intercepts DE and EF on q also.**

**Ans. **In fig are 3 parallel lines intersected by two transversal P and Q.

To Prove DE=EF

Proof: In

B is mid-point of AC

And BG||CF

G is mid-point of AF [By mid-point theorem]

Now In

G is mid-point of AF and GE || AD

E is mid-point of FD [By mid-point theorem]

DE=EF

Hence Proved.

**21. ABCD is a parallelogram in which E is mid-point of AD. DF||EB meeting AB produced at F and BC at L prove that DF = 2DL**

**Ans. **In

is mid-point of AD (Given)

BE||DF (Given)

By converse of mid-point theorem B is mid-point of AF

ABCD is parallelogram

From (i) and (ii)

CD = BF

Consider and

DC = FB [Proved above]

[Alternate angles]

[Vertically opposite angles]

[ASA]

DL=LF

DF=2DL

**22. PQRS is a rhombus if find **

**Ans. **[opposite angles of a parallelogram are equal]

Let

In we have RS=RQ

[ opposite Sides of equal angles are equal]

In

[By angle sum property]

**23. ABCD is a trapezium in which AB||CD and AD = BC show that**

**(i) **

**(ii) **

**(iii) **

**Ans. **Produce AB and Draw a line Parallel to DA meeting at E

AD||EC

….(i) [Sum of interior angles on the some side of transversal is ]

In

BC=CE (given)

…..(2) [in a equal side to opposite angles are equal]

……(3)

By (i) and (3)

**(i) **

**(ii) **

**(iii) **In ABC and BAD

AB=AB [common]

[Proved above]

AD=BC [given]

[By SAS]

**24. Show that diagonals of a rhombus are perpendicular to each other.**

**Ans. **Given: A rhombus ABCD whose diagonals AC and BD intersect at a Point O

To Prove:

Proof: clearly ABCD is a Parallelogram in which

AB=BC=CD=DA

We know that diagonals of a Parallelogram bisect each other

OA=OC and OB=OD

Now in BOC and DOC, we have

OB=OD

BC=DC

OC=OC

BOC DOC [By SSS]

[By C.P.C.T]

But

Similarly,

Hence diagonals of a rhombus bisect each other at

**25. Prove that the diagonals of a rhombus bisect each other at right angles**

**Ans. **We are given a rhombus ABCD whose diagonals AC and BD intersect each other at O.

We need to prove that OA=OC, OB=OD and

In and

AB=CD [Sides of rhombus]

[vertically opposite angles]

And [Alternate angles]

AOBCOD [By ASA]

OA=OC

And OB=OD [By C.P.C.T]

Also in AOB and COB

OA=OC [Proved]

AB=CB [sides of rhombus]

And OB=OB [Common]

AOB COB [By SSS]

[By C.P.C.T]

But [linear pair]

**26. In fig ABCD is a trapezium in which AB||DC and AD=BC. Show that **

**Ans. **To show that

Draw CP||DA meeting AB at P

AP||DC and CP||DA

APCD is a parallelogram

Again in CPB

CP=CB [BC=AD [Given]

[Angles opposite to equal sides]

But [By linear pair]

Also [APCD is a parallelogram]

Or

= CB

**27. In fig ABCD and ABEF are Parallelogram, prove that CDFE is also a parallelogram.**

**Ans. **ABCD is a parallelogram

AB=DC also AB||DC…………..(i)

Also ABEF is a parallelogram

AB=FE and AB||FE……….(ii)

By (i) and (ii)

AB=DC=FE

AB=FE

And AB||DC||FE

AB||FE

CDEF is a parallelogram.

Hence Proved.

**4 Marks Quetions**

**1. 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 both B as well as D.**

**Ans**. ABCD is a rectangle. Therefore AB = DC ……….(i)

And BC = AD

Also A = B = C = D =

**(i)** In ABC and ADC

1 = 2 and 3 = 4

[AC bisects A and C (given)]

AC = AC [Common]

ABC ADC [By ASA congruency]

AB = AD ……….(ii)

From eq. (i) and (ii), AB = BC = CD = AD

Hence ABCD is a square.

**(ii)** In ABC and ADC

AB = BA [Since ABCD is a square]

AD = DC [Since ABCD is a square]

BD = BD [Common]

ABD CBD [By SSS congruency]

ABD = CBD [By C.P.C.T.] ……….(iii)

And ADB = CDB [By C.P.C.T.] ……….(iv)

From eq. (iii) and (iv), it is clear that diagonal BD bisects both B and D.

**2. An ABC and DEF, AB = DE, AB DE, BC = EF and BC EF. Vertices A, B and C are joined to vertices D, E and F respectively (See figure). 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**

**Ans. (i) **In ABC and DEF

AB = DE [Given]

And AB DE [Given]

ABED is a parallelogram.

**(ii) **In ABC and DEF

BC = EF [Given]

And BC EF [Given]

BEFC is a parallelogram.

**(iii) **As ABED is a parallelogram.

AD BE and AD = BE ……….(i)

Also BEFC is a parallelogram.

CF BE and CF = BE ……….(ii)

From (i) and (ii), we get

AD CF and AD = CF

**(iv)** As AD CF and AD = CF

ACFD is a parallelogram.

**(v)** As ACFD is a parallelogram.

AC = DF

**(vi)** InABC and DEF,

AB = DE [Given]

BC = EF [Given]

AC = DF [Proved]

ABC DEF [By SSS congruency]

**3. 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 figure). Show that F is the mid-point of BC.**

**Ans**. Let diagonal BD intersect line EF at point P.

In DAB,

E is the mid-point of AD and EP AB [EF AB (given) P is the part of EF]

P is the mid-point of other side, BD of DAB.

[A line drawn through the mid-point of one side of a triangle, parallel to another side intersects the third side at the mid-point]

Now in BCD,

P is the mid-point of BD and PF DC [EF AB (given) and AB DC (given)]

EF DC and PF is a part of EF.

F is the mid-point of other side, BC of BCD. [Converse of mid-point of theorem]

**4. Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.**

**Ans. **Given: A quadrilateral ABCD in which EG and FH are the line-segments joining the mid-points of opposite sides of a quadrilateral.

To prove: EG and FH bisect each other.

Construction: Join AC, EF, FG, GH and HE.

Proof: In ABC, E and F are the mid-points of respective sides AB and BC.

EF AC and EF AC ……….(i)

Similarly, in ADC,

G and H are the mid-points of respective sides CD and AD.

HG AC and HG AC ……….(ii)

From eq. (i) and (ii),

EF HG and EF = HG

EFGH is a parallelogram.

Since the diagonals of a parallelogram bisect each other, therefore line segments (i.e. diagonals) EG and FH (of parallelogram EFGH) bisect each other.

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

**Ans. **Let ABCD be a quadrilateral in which equal diagonals AC and BD bisect each other at right angle at point O.

We have AC = BD and OA = OC ……….(i)

And OB = OD ……….(ii)

Now OA + OC = OB + OD

OC + OC = OB + OB [Using (i) & (ii)]

2OC = 2OB

OC = OB ……….(iii)

From eq. (i), (ii) and (iii), we get, OA = OB = OC = OD ……….(iv)

Now in AOB and COD,

OA = OD [proved]

AOB = COD [vertically opposite angles]

OB = OC [proved]

AOBDOC [By SAS congruency]

AB = DC [By C.P.C.T.] ……….(v)

Similarly, BOC AOD [By SAS congruency]

BC = AD [By C.P.C.T.] ……….(vi)

From eq. (v) and (vi), it is concluded that ABCD is a parallelogram because opposite sides of a quadrilateral are equal.

Now in ABC and BAD,

AB = BA [Common]

BC = AD [proved above]

AC = BD [Given]

ABC BAD [By SSS congruency]

ABC = BAD [By C.P.C.T.] ……….(vii)

But ABC + BAD = [ABCD is a parallelogram] ……….(viii)

AD BC and AB is a transversal.

ABC + ABC = [Using eq. (vii) and (viii)]

2ABC = ABC =

ABC = BAD = ……….(ix)

Opposite angles of a parallelogram are equal.

But ABC = BAD =

ABC = ADC = ……….(x)

BAD = BDC = ……….(xi)

From eq. (x) and (xi), we get

ABC = ADC = BAD = BDC = ……….(xii)

Now in AOB and BOC,

OA = OC [Given]

AOB = BOC = [Given]

OB = OB [Common]

AOBCOB [By SAS congruency]

AB = BC ……….(xiii)

From eq. (v), (vi) and (xiii), we get,

AB = BC = CD = AD ……….(xiv)

Now, from eq. (xii) and (xiv), we have a quadrilateral whose equal diagonals bisect each other at right angle.

Also sides are equal make an angle of with each other.

ABCD is a square.

**6. ABCD is a trapezium in which AB **** CD and AD = BC (See figure). Show that:**

**(i) ****A = ****B**

**(ii) ****C = ****D**

**(iii) ****ABC ****BAD**

**(iv) Diagonal AC = Diagonal BD **

**Ans. **Given: ABCD is a trapezium.

AB CD and AD = BC

To prove: **(i)** A = B

**(ii)** C = D

**(iii)** ABC BAD

**(iv)** Diag. AC = Diag. BD

Construction: Draw CE AD and extend AB to intersect CE at E.

Proof: (i) As AECD is a parallelogram. [By construction]

AD = EC

But AD = BC [Given]

BC = EC

3 = 4 [Angles opposite to equal sides are equal]

Now 1 + 4 = [Interior angles]

And 2 + 3 = [Linear pair]

1 + 4 = 2 + 3

1 = 2 [3 = 4 ]

A = B

**(ii) **3 = C [Alternate interior angles]

And D = 4 [Opposite angles of a parallelogram]

But 3 = 4 [BCE is an isosceles triangle]

C = D

**(iii)** In ABC and BAD,

AB = AB [Common]

1 = 2 [Proved]

AD = BC [Given]

ABC BAD [By SAS congruency]

AC = BD [By C.P.C.T.]

**7. Prove that if the diagonals of a quadrilateral are equal and bisect each other at right angles then it is a square.**

**Ans. **Given in a quadrilateral ABCD, AC = BD, AO = OC and BO = OD and

To prove: ABCD is a square.

Proof:

OA=OC

OB=OD [given]

And

[vertically opposite angles]

But these are alternate angles

ABCD is a parallelogram whose diagonals bisects each other at right angles

Again in

AB=BC [Sides of a rhombus]

AD=AB [Sides of a rhombus]

And BD=CA [Given]

[By CPCT]

These are alternate angles of these same side of transversal

Hence ABCD is a square.

**8. Prove that in a triangle, the line segment joining the mid points of any two sides is parallel to the third side.**

**Ans. **Given: A in which D and E are mid-points of the side AB and AC respectively

__To Prove:__

__Construction:__ Draw

__Proof:__ In and

[Vertically opposite angles]

AE=CE [Given]

And [Alternate interior angles]

[By ASA]

DE=FE [By C.P.C.T]

But DA = DB

DB = FC

Now DBFC

DBCF is a parallelogram

DEBC

Also DE = EF = BC

**9. 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 quadrilateral PQRS is a rectangle. **

**Ans. **Join AC and BD which intersect at O let BD intersect RS at E and AC intersect RQ at F

IN ABD P and S are mid-points of sides AB and AD.

PS||BD and PS=

Similarly, RQ||DB and RQ=BD

RS||BD||RQ and PS =

PS=RQ and PS||RQ

PQRS is a parallelogram

Now RF||EO and RE||FO

OFRE is also a parallelogram.

Again, we know that diagonals of a rhombus bisect each other at right angles.

[opposite angles of a parallelogram]

Each angle of the parallelogram PQRS is

Hence PQRS is a rectangle.

**10. In the given Fig ABCD is a parallelogram E is mid-point of AB and CE bisects Prove that: **

**(i) AE = AD **

**(ii) DE bisects **

**(iii) **

**Ans. **ABCD is a parallelogram

And EC cuts them

[Alternate interior angle]

**(i) **Now AE=AD

[ Alternate interior angles]

**(ii) ** DE bisects

**(iii) **Now

But, the sum of all the angles of the triangle is

**11. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. show that (i) D is mid-point of AC (ii) MDAC (iii) CM = MA = AB**

**Ans**. Given ABC is a right angle

**at C**

**(i) **M is mid-point of AB

And MD||BC

D is mid-Point of AC [a line through midpoint of one side of a parallel to another side bisect the third side.

**(ii).** MD||BC

[Corresponding angles]

**(iii) **In ADM and CDM

AD=DC [D is mid-point of AC]

DM=DM [Common]

ADM CDM [By SAS]

AM=CM [By C.P.C.T]

AM=CM=MB [ M is mid-point of AB]

CM=MA=AB.