Important Questions for CBSE Class 9 Maths Chapter 8 – Quadrilaterals

1 Marks Quetions

1. A quadrilateral ABCD is a parallelogram if

(a) AB = CD

(b) ABBC

(c) 

(d) AB = AD

Ans. (c) 

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

(a) 

(b) 

(c) 

(d) 

Ans. (c) 

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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

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4. ABCD is a rhombus. If 

(a) 

(b) 

(c) 

(d) 

Ans. (c) 

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

Ans. 

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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

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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

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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

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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

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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

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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)

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12. ABCD is rhombus with  The measure of is

(a) 

(b) 

(c) 

(d) 

Ans. b) 

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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

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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

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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

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

(a) Kite

(b) trapezium

(c) cyclic quadrilateral

(d) parallelogram

Ans. (d) parallelogram

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

(a) and

Ans. In 

$���������������������������������$

 $�������������$

Now 

Also CD||BA Given)

is a parallelogram

(ii) ABCD is a parallelogram

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18. Which of the following is not a parallelogram?

(a) Rhombus

(b) Square

(c) Trapezium

(d) Rectangle

Ans. (c) Trapezium

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

(a) 1800

(b) 3600

(c) 2700

(d) 900

Ans. (b) 3600

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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

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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

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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 

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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 $������$

AC = BD $�����$

AD = BC [opp. Sides of a gm]

 ABC BAD $���������������$

 DAB = CBA $���.�.�.�.$ ……….(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.

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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 $���������������$ ……….(i)
Similarly 2 = 4 ……….(ii)
But 1 = 2 $�����$ ……….(iii)
 3 = 4 $�������.(�),(��)���(���)$
Thus AC bisects C.

(ii) 2 = 3 = 4 = 1
 AD = CD $��������������������������$
 AB = CD = AD = BC
Hence ABCD is a rhombus.

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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 $�����������������������$

AB = CD $�������������������������������������$

APB = CQD = 

 APB CQD $���������������$

(ii) Since APB CQD

AP = CQ $���.�.�.�.$

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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 $�������������������������������������������ℎ��ℎ��$

Now PQ = SR and PQ  SR

Therefore, PQRS is a parallelogram.

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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 ]

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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

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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.

$�����������������������$

$���������$

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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 $���������������������������������$

AB = AB common $������$

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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 $������$

is a parallelogram.

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11. 
Ans. ABCD is a parallelogram. The diagonals of a parallelogram bisect bisect each other

But $�����$

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.

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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.

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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 

$���������ℎ�����$

$���������ℎ�����$

$������$

 $���������������$

And 

Hence AC bisects as well as 

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

Hence BD bisects as well as 

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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.

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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 

 $�����$

 $�����$

And [Vertically apposite angle

 $�����$

$���.�.�.�$

But this is Pair of alternate interior angles

Similarly AD||BC

Quadrilateral ABCD is a Parallelogram.

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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 $�����������������������������$

And 

$���$

(II) (By C.P.C.T)

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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.

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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: $�����$

$�����$

In quadrilateral. ABCD

 $��\dots .(�)$

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

and 

ABCD is a parallelogram.

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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

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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 

$�����������-��������������������������$

 is a parallelogram

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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

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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.

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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.

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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.

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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$�����$

AOD = BOC$�����$

OB = OD$�����$

AODCOB $���������������$

AD = CB $���.�.�.�.$……….(i)

Again, In AOB and COD,

OA = OC$�����$

AOB = COD$�����$

OB = OD$�����$

AOBCOD $���������������$

AD = CB$���.�.�.�.$……….(ii)

Now In AOD and BOC,

OA = OC$�����$

AOB = BOC$�����$

OB = OB$������$

AOBCOB $���������������$

AB = BC $���.�.�.�.$……….(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.

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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$������$

ABC = BAD = 

BC = AD $��������������$

ABC BAD $���������������$

AC = BD $���.�.�.�.$ Hence proved.

Now in triangles AOB and AOD,

AO = AO$������$

AB = AD$��������������$

OB = OD$���������������������������ℎ��ℎ��$

AOBAOD$���������������$

AOB = AOD$���.�.�.�.$

ButAOB + AOD = $����������$

AOB = AOD = 

OA  BD or AC  BD

Hence proved.

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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$������$

AB = AD$�������������ℎ�����$

OB = OD(diagonals of rhombus bisect each other]

AOBAOD$���������������$

OAD = OAB$���.�.�.�.$

OA bisects A……….(i)

SimilarlyBOC DOC$���������������$

OCB = OCD$���.�.�.�.$

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$������$

AB = BC$�������������ℎ�����$

OA = OC(diagonals of rhombus bisect each other]

AOBCOB$���������������$

OBA = OBC$���.�.�.�.$

OB bisects B……….(iii)

SimilarlyAODCOD$���������������$

ODA = ODC$���.�.�.�.$

BD bisects D……….(iv)

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

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

(i) APDCQB

(ii) AP = CQ

(iii) AQBCPD

(iv) AQ = CP

(v) APCQ is a parallelogram.

Ans. (i)In APD and CQB,

DP = BQ$�����$

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

AD = CB$����������������������������$

APDCQB$���������������$

(ii) SinceAPDCQB

AP = CQ$���.�.�.�.$

(iii) In AQB and CPD,

BQ = DP$�����$

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

AB = CD$����������������������������$

AQBCPD$���������������$

(iv) SinceAQBCPD

AQ = CP$���.�.�.�.$

(v) In quadrilateral APCQ,

AP = CQ$������������(�)$

AQ = CP$������������(��)$

Since opposite sides of quadrilateral APCQ are equal.

Hence APCQ is a parallelogram.

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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. $�����$

AB = BC

 AB =  BCPB = BQ

1 = 2$����������������������������������$

Now in triangles APS and CQR, we have,

AP = CQ$���������ℎ����-��������������������=��$

Similarly AS = CR and PS = QR$�����������������������������$

APS CQR$����������������$

3 = 4$���.�.�.�.$

Now we have1 + SPQ + 3 = 

And2 + PQR + 4 = $�����������$

1 + SPQ + 3 = 2 + PQR + 4

Since1 = 2 and 3 = 4$�����������$

SPQ = PQR……….(iii)

Now PQRS is a parallelogram $�����������$

SPQ + PQR = ……….(iv)$��������������$

Using eq. (iii) and (iv),

SPQ + SPQ = 2SPQ = 

SPQ = 

Hence PQRS is a rectangle.

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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.$�����$

AD = BC

AD = BCAS = BQ……….(iv)

In triangles APS and BPQ,

AP = BP$����ℎ����-���������$

PAS = PBQ[Each  ]

And AS = BQ$������.(��)$

 APS BPQ$���������������$

PS = PQ$���.�.�.�.$………(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.

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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$������.(�)$

AECF is a parallelogram.

FA  CEFP CQ$�����������������������������$ ……...(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.

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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$�����$

MD  BC

AD = DC$�������������-������ℎ�����$

Thus D is the mid-point of AC.

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

1 = C$�������������������$

1 = [C = ]

Thus MD  AC.

(iii)  In AMD and CMD,

AD = DC$�����������$

1 = 2 = $�����������$

MD = MD$������$

AMD CMD$���������������$

AM = CM$���.�.�.�.$……….(i)

Given that M is the mid-point of AB.

AM =  AB……….(ii)

From eq. (i) and (ii),

CM = AM =  AB

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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

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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 $������$

 $�����$

$������$

$������$

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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) $�����������������������������$

$�����$ …(ii)

And $����$ …..(iii)

But 

By (2) and (3), we get

Also, $����������������������������$ ….(v)

 From (i), (iv) and (v), we get

But, AB =CD $����������������������������$

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
$������������������������$
EF=FD $��������������$

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

$������$

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 

$���������������$

$���������������$

And $�����$

$���$

$���.�.�.�$

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$��+��+��$

= 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

$����������������������������������������������ℎ����������������������$

(ii) Similarly, quadrilateral DPBQ is a Parallelogram because DQ||PB and DQ=PB

(iii) In quadrilateral PSQR,

SP||QR $�����������������������������$

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 $�����-������ℎ�����$

Now In 

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

E is mid-point of FD $�����-������ℎ�����$ 

 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 $�����������$

$���������������$

$������������������������$

$���$

DL=LF

DF=2DL

---

22. PQRS is a rhombus if find 

Ans. $��������������������������������������$

Let 

In we have RS=RQ

$����������������������������������$

In 

$������������������$

---

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 $������$

$�����������$

AD=BC $�����$

$�����$

---

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 $�����$

$���.�.�.�$

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 $��������ℎ�����$

 $������������������������$

And  $���������������$

AOBCOD $�����$

 OA=OC

And OB=OD $���.�.�.�$

Also in AOB and COB

OA=OC $������$

AB=CB $��������ℎ�����$

And OB=OB $������$

AOB COB $�����$

$���.�.�.�$

But $����������$

---

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 $�����$

 $��������������������������$

But $������������$

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 $������$

 ABC ADC $���������������$

 AB = AD ……….(ii)

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

Hence ABCD is a square.

(ii) In ABC and ADC

AB = BA $������������������$

AD = DC $������������������$

BD = BD $������$

 ABD CBD $���������������$

 ABD = CBD $���.�.�.�.$ ……….(iii)

And ADB = CDB $���.�.�.�.$ ……….(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 $�����$

And AB  DE $�����$

 ABED is a parallelogram.

(ii) In ABC and DEF

BC = EF $�����$

And BC EF $�����$

 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 $�����$

BC = EF $�����$

AC = DF $������$

 ABC DEF $���������������$

---

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.

$�����������ℎ����ℎ�ℎ����-�������������������������,��������������ℎ�����������������ℎ��ℎ����������ℎ����-�����$

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. $�������������-��������ℎ�����$

---

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 

 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 $������$

AOB = COD $������������������������$

OB = OC $������$

 AOBDOC $���������������$

 AB = DC $���.�.�.�.$ ……….(v)

Similarly, BOC AOD $���������������$

 BC = AD $���.�.�.�.$ ……….(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 $������$

BC = AD $�����������$

AC = BD $�����$

 ABC BAD $���������������$

 ABC = BAD $���.�.�.�.$ ……….(vii)

But ABC + BAD =  $��������������������$ ……….(viii)

 AD  BC and AB is a transversal.

 ABC + ABC =  $�������.(���)���(����)$

 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 $�����$

AOB = BOC =  $�����$

OB = OB $������$

 AOBCOB $���������������$

 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. $��������������$

 AD = EC

But AD = BC $�����$

 BC = EC

 3 = 4 $����������������������������������$

Now 1 + 4 = $��������������$

And 2 + 3 = $����������$

 1 + 4 = 2 + 3

 1 = 2 [3 = 4 ]

 A = B

(ii) 3 = C $�����������������������$

And D = 4 $������������������������������$

But 3 = 4 [BCE is an isosceles triangle]

 C = D

(iii) In ABC and BAD,

AB = AB $������$

1 = 2 $������$

AD = BC $�����$

 ABC BAD $���������������$

 AC = BD $���.�.�.�.$

---

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 $�����$

And

$������������������������$

But these are alternate angles 

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

Again in 

AB=BC $���������ℎ�����$

AD=AB $���������ℎ�����$

And BD=CA $�����$

$������$

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 

 $������������������������$

AE=CE $�����$

And  $�����������������������$

 $�����$

DE=FE $���.�.�.�$

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.

$������������������������������$

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

 $����������������������$

(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

 $�������������������$

(iii) In  ADM and CDM

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

DM=DM $������$

ADM CDM $�����$

AM=CM $���.�.�.�$

AM=CM=MB [ M is mid-point of AB]

 CM=MA=AB.