Quadrilaterals

1. A quadrilateral ABCD is a parallelogram if

(a) AB = CD (b) AB �BC

(c) 0 0 060 , 60 , 120A C B∠ = ∠ = ∠ = (d) AB = AD

2. In figure, ABCD and AEFG are both parallelogram if 080 , C then DGF is∠ = ∠

(a) 0100

(b) 060

(c) 080

(d) 0120

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

(a) acute angled (b) obtuse angled

(c) equilateral (d) right angled

4. ABCD is a rhombus. If 030 , ACB then ADB is∠ = ∠

(a) 030 (b) 0120

(c) 060 (d) 045

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

6. Show that each angles of a rectangle is a right angle.

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

8. Prove that diagonals of a rectangle are equal in length.

9. In a parallelogram ABCD, bisectors of adjacent angles A and B intersect each other at P. prove that 090APB∠ =

10. In figure diagonal AC of parallelogram ABCD bisects A∠ show that (i) if bisects C∠ and (ii) ABCD is a rhombus

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

parallelogram.

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

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

ANSWERS

Ans01. (c) Ans02. (c) Ans03. (d) Ans04. (c)

Ans05. Suppose angles of quadrilateral ABCD are 3x, 5x, 9x, and 13x

0 0

0

0

0

0

360 [ 360 ]

3 5 9 13 360

30 360

12

3 3 12 36

A B C D sum of angles of a quadrilateral is

x x x x

x

x

A x

∠ + ∠ + ∠ + ∠ =+ + + =

==

∴ ∠ = = × =0

0

0

5 5 12 60

9 9 12 108

13 13 12 156

B x

C x

D x

∠ = = × =∠ = = × =∠ = = × =

Ans06. We know that rectangle is a parallelogram whose one angle is right angle.

Let ABCD be a rectangle. 090A∠ =

To prove 090B C D∠ = ∠ = ∠ = Proof: AD BC∵ � and AB is transversal

0

0 0

0 0

0

0

0

180

90 180

180 90

90

90

90

A B

B

B

C A

C

D B

D

∴ ∠ + ∠ =+ ∠ =

∠ = −=

∠ = ∠∴ ∠ =

∠ = ∠∴ ∠ =

Ans07. AB CD∵ � and EF cuts them at P and R.

[ int ]

1 1

2 2. . 1 2

[ ]

APR PRD Alternate erior angles

APR PRD

i e

PQ RS Alternate

∴ ∠ = ∠

∴ ∠ = ∠

∠ = ∠∴ �

Ans08. ABCD is a rectangle AC and BD is diagonals.

To prove AC = BD

Proof: In DAB and CBA∆ AD = BC [In a rectangle opposite sides are equal]

0 [90 ]A B each∠ = ∠ AB = AB common [common]

[ ]

[ ]

DAB CAB By SAS

AC BD By CPCT

∴ ∆ ≅ ∆∴ =

Ans09. Given ABCD is a parallelogram is and bisectors of A and B∠ ∠ intersect each

other at P.

To prove 090APB∠ =Proof:

( )

1 11 2

2 21

( )2

A B

A B i

∠ + ∠ = ∠ + ∠

= ∠ + ∠ →

But ABCD is a parallelogram AD �BC

0

0

0

0

180

1 1 2 180

290

1 2 180 [ ]A B

In APB

APB By angle sum property∴ ∠ + ∠ =

∴ ∠ + ∠ = ×

=

∆∠ + ∠ + ∠ =

0 0

0

90 180

90

APB

APB

Hence proved

+ ∠ =∠ =

Ans10. (i) AB�DC and AC is transversal

1 2 ( )

3 4 ( )

1 3

2 4

sec

Alternate angles

and Alternate angles

But

AC bi sts C

∴ ∠ = ∠∠ = ∠∠ = ∠

∴ ∠ = ∠∴ ∠

(ii) In ABC and ADC∆ ∆ [ ]

1 3 [ ]

2 4 [ ]

[ ]

AC AC common

given

proved

ABC ADC

AB AD By CPCT

=∠ = ∠

∠ = ∠∴ ∆ ≅ ∆∴ =∴ homABCD is a r bus

Ans11. Given in a parallelogram AX and CY bisects A and C∠ ∠ respectively and we have

to show that AYCX in a parallelogram.

( ) [ log ]

1 [ ] ( )

21

BCY= C 2

In ADX and CBY

D B i opposite angles of a paralle ram

DAX A given ii

and

∆ ∆∠ = ∠ →

∠ = ∠ →

∠ ∠ [given] (iii)

But A= C

By (2) and (3), we get

→

∠ ∠∴

( )

[ log ] ( )

( ), ( ) ( ),

[ ]

DAX BCY iv

Also AD BC opposite sides of paralle ram v

From i iv and v we get

ADX CBY By ASA

DX BY

∠ = ∠ →= →

∴∆ ≅ ∆

∴ = [ ]

[ log ]

[ ]

log

CPCT

But AB CD oppsite sides of paralle ram

AB BY CD DX

or

AY CX

But AY XC ABCD is a gm

AYCX is a paralle ram

=− = −

=

∴� ∵ �

Ans12. 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: In AEF and CDF∆ ∆ [ int ]

1 2 [ ]

[ ]

[ ]

AF FC F is mid po of AC

vertically opposite angles

EF FD BY construction

AEF CDF By SAS

= −∠ = ∠

=∴ ∆ ≅ ∆∴

∵

[ ]

[ int]

[ ]

log

AE CD By CPCT

and AE BE E is the mid po

BE cd

and AB CD BAC ACD

BCDE is a paralle ram

EF BC Hencep

== −

∴ =∴∠ = ∠

∴

∵

�

� roved

Ans13. Given in a quadrilateral ABCD, AC = BD, AO = OC and BO = OD and 090AOB∠ =To prove: ABCD is a square.

Proof: In AOB and COD∆ ∆

[ ]

[ ]

[ ]

[ ]

[

OA OC given

OB OD given

and AOB COD vertically opposite angles

AOB COD By SAS

AB CD By CPC

==

∠ = ∠∴ ∆ ≅ ∆∴ = ]

1 2 [ ]

is a parallelogram whose diagonals bisects each other at right angles

rhombus

T

By CPCT

But there are alternate angles AB CD

ABCD

ABCD is a

Again in ABD and

∠ = ∠∴

∴∆ ∆

�

[ hom ]

[ hom ]

[ ]

[

BCA

AB BC Sides of a r bus

AD AB Sides of a r bus

and BD CA Given

ABD BCA

BAD CBA By CPCT

===

∴ ∆ ≅ ∆∴ ∠ = ∠

0 0

]

180 90

There are alternate angles of these same side of transversal

BAD CBA or BAD CBA

Hence ABCD is a square

∴ ∠ + ∠ = ∠ = ∠ =

1. In fig ABCD is a parallelogram. It 60DAB∠ = ° and 80DBC∠ = ° then CDB∠ is

( )( )( )( )

A 80°

B 60°

C 20°

D 40°

2. If the diagonals of a quadrilateral bisect each other, then the quadrilateral must

be.

(a) Square

(b) Parallelogram

(c) Rhombus

(d) Rut angle

A B

CD

600

800

Quadrilateral

3. 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) Rut angle

4. 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

5. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

6. 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.

AB

CD

Y

X

O

7. Show that the line segments joining the mid points of opposite sides of a quadrilateral

bisect each other.

A B

CD

E

F

G

H

8. ABCD is a rhombus show that diagonal AC bisects A∠ as well as C∠ and diagonal BD bisects B∠ as well as D∠

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

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

11. In fig B∠ is a right angle in .ABC D∆ is the mid-point of .AC DE AB� intersects BC at E. show that

(i) E is the mid-point of BC

(ii) DE ⊥ BC

(iii) BD = AD

A

BC

D

E

12. 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 DEF∆ is double the perimeter of ABC∆ .

13. Prove that in a triangle, the line segment joining the mid points of any two sides isparallel to the third side.

ANSWERS

Ans1. (D) Ans2. (B) Ans3. (B) Ans4. (A)

Ans5. Given A quadrilateral ABCD in which AB = DC and AD = BC

To prove: ABCD is a parallelogram

Construction: Join AC

Prof: In ABC∆ and ADC∆

( )

[ ]

given

common

===

AD BC

AB DC

AC AC

[ ][ ]by sss

By CPCT

ABC ADC

BAC DAC

∴∆ ≅ ∆

∴∠ = ∠

ABCD∴ Is a parallelogram

B

CD

A

Ans6. ABCD is a parallelogram. The diagonals of a parallelogram bisect bisect each

other

OD OB∴ =

But DX BY= [given]

OD DX OB BY

or OX OY

∴ − = −=

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

AYCX∴ is a parallelogram.

Ans7. 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 ABC∆ , E is mid-point of AB and F is mid-point of BC

EF AC∴ � And ( )1.......

2EF AC i=

Similarly HG AC� and ( )1......

2GD AC ii=

From (i) and (ii), EF HG� and EF GH=

EFGH∴ is a parallelogram and EG and HF are its diagonals

Diagonals of a parallelogram bisect each other

Thus EG and HF bisect each other.

Ans8. ABCD is a rhombus

In ABC∆ and ADC∆

AB AD= [Sides of a rhombus]

BC DC= [Sides of a rhombus]

AC AC= [Common]

ABC ADC∴∆ ≅ ∆ [By SSS Congruency]

CAB CAD∴∠ = ∠ And ACB ACD∠ = ∠

Hence AC bisects A∠ as well as C∠

Similarly, by joining B to D, we can prove that ABD CBD∆ ≅ ∆

Hence BD bisects B∠ as well as D∠

B

CD

A

Ans9. 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

,OA OC OB OD∴ = = And 1 2 3 90∠ = ∠ = ∠ = °

Now In BOA∆ And BOC∆

OA OC= Given

OB OB= [Common]

B

CD

A

0

1

23

And 1 2 90∠ = ∠ = ° (Given)

BOA BOC∴∆ = ∆ (SAS)

BA BC∴ = (C.P.C.T.)

Ans10. Given a trapezium ABCD in which AB DC� and M,N are the mid Points of the

diagonals AC and BD.

We need to prove that MN AB DC� �

Join CN and let it meet AB at E

Now in CDN∆ and ∆EBN

DCN BEN∠ = ∠ [Alternate angles]

CDN BEN∠ = ∠ [Alternate angles]

And DN BN= [given]

CDN EBN∴∆ ≅ ∆ [SAA]

CN EN∴ = [By C.P.C.T]

Now in ,ACE M∆ and N are the mid points of the sides AC and CE respectively.

MN AE∴ � Or MN AB�

Also AB DC�

MN AB DC∴ � �

EB

CD

A

M N

Ans11. Proof: DE AB∵ � and D is mid points of AC

In DCE∆ and DBE∆

CE=BE

DE= DE

And 90DEC DEB∠ = ∠ = °

DCE DBE

DCE DBE

CD BD

∴∆ = ∆∴∆ ≅ ∆∴ =

Ans12. ∵BCAF Is a parallelogram

BC AF∴ =

ABCE∵ Is a parallelogram

2

BC AE

AF AE BC

∴ =+ =

Or 2EF BC=

Similarly ED = 2AB and FD = 2AC

∴Perimeter of ABC AB BC AC∆ = + +

Perimeter of DEF DE EF DF∆ = + +

= 2AB+2BC+2AC

= 2[AB+BC+AC]

= 2 Perimeter of ABC∆

Hence Proved

E

B C

D

A

F

Ans13. Given: A ABC∆ in which D and E are mid-points of the side AB and AC

respectively

TO Prove: DE BC�

Construction: Draw CF BA�

Proof: In ADE∆ and CFE∆

1 2∠ = ∠ [Vertically opposite

angles]

AE=CE [Given]

And 3 4∠ = ∠ [Alternate interior angles]

ADE CFE∴∆ ≅ ∆ [By ASA]

∴DE=FE [By C.P.C.T]

But DA = DB

∴DB = FC

Now DB� FC

∴DBCF is a parallelogram

∴DE�BC

Also DE = EF = 1

2BC

E

BC

D

A

3

1

2

4

F