Quadrilaterals

QUADRILATERAL :QUADRILATERAL :

A quadrilateral is a geometrical figure which has four sides, four angles, four vertices, and two diagonals. The sum of all angles of a quadrilateral is 360

A B

C D

A B

C D

There are actually six types of quadrilaterals. They are as follows: Trapezium

Parallelogram

Rectangle

Rhombus

Square

Kite

TRAPEZIUM :

If in a quadrilateral one pair of opposite sides are equal then the given quadrilateral is called a TRAPEZIUM.TRAPEZIUM.

E.g. : In the above figure if AB is parallel to CD then the figure is a quadrilateral.

A B

CD

PARALLELOGRAM:

If in a quadrilateral both the pairs of opposite sides are parallel , then the given quadrilateral is a PARALLELOGRAM.PARALLELOGRAM.

E.G.- In the above figure if AB is parallel to CD and AD is parallel to BC then the figure is a parallelogram.

A B

C D

RECTANGLE:

If in a quadrilateral one of its angles is a right angle then the quadrilateral is a rectangle.

E.G. : If in the above figure angle A is a right angle then the figure is a rectangle.

A B

CD

A B

D C

RHOMBUS:

If in a parallelogram all sides are equal, then the parallelogram is a RHOMBUS.

E.G. : If in the above figure AB=BC=CD=DA, then it is rhombus.

SQUARE:

A parallelogram whose one angle is a right angle and all the sides are equal, then it is called a SQUARE.

E.G.: If in the above figure AB=BC=CD=DA, and angle B is a right angle, then the given figure is a square.

A B

CD

A

B

C

D

KITE:KITE:

In a quadrilateral if two pairs of adjacent sides are equal. Then it is not a parallelogram. It is called a KITE.

E.G. : If in the above figure AB = AD and BC = CD, then it is not a parallelogram. It is a kite.

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.

A rectangle or a rhombus is not a square .

The sum of angles of a quadrilateral is 360 degrees.

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

In a parallelogram opposite sides are equal.

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

In a parallelogram opposite angles are equal.

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

The diagonals of a parallelogram bisect each other.

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

A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

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

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

Theorem : Sum of angles of a quadrilateral is 360

A B

CD

Given: A quadrilateral ABCD.

To prove: angles A + B+ C+ D= 360.

Construction: Join A to C.

Proof: In triangle ABC,

angle CAB + angle ACB + angle CBA = 180. (A.S.P) – 1

In triangle ACD,

angle ADC + angle DCA + angle CAB = 180 (A.S.P) -2

Adding 1 and 2

angles CAB+ACB+CBA+ADC+DCA+CAD=180+180

angles (CAB+BAC)+ABC+(BCA+ACD)+ADC= 360.

Therefore, angles A+B+C+D=360.

THEOREM: The diagonal of a parallelogram divides it into two congruent triangles.

A B

CD

Given: A parallelogram ABCD and its diagonal AC.

To prove: Triangle ABC is congruent to triangle ADC

Construction: Join A to C.

Proof: In triangles ABC and ADC,

AB is parallel to CD and AC is the transversal

Angle BAC = Angle DCA (alternate angles)

Angle BCA = Angle DAC (alternate angles)

AC = AC (common side)

Therefore, triangle ABC is congruent to triangle ADC by ASA rule.

Hence Proved.

Given: A parallelogram ABCD and its diagonal AC.

To prove: Triangle ABC is congruent to triangle ADC

Construction: Join A to C.

Proof: In triangles ABC and ADC,

AB is parallel to CD and AC is the transversal

Angle BAC = Angle DCA (alternate angles)

Angle BCA = Angle DAC (alternate angles)

AC = AC (common side)

Therefore, triangle ABC is congruent to triangle ADC by ASA rule.

Hence Proved.

THEOREM: In a parallelogram , opposite sides are equal. D

A B

CGiven: A parallelogram ABCD.

To Prove: AB = DC and AD = BC

Construction: Join A to C

Proof: In triangles ABC and ADC,

AB is parallel to CD and AC is the transversal.

Angle BAC = Angle DCA (alternate angles)

Angle BCA = Angle DAC (alternate angles)

AC = AC (common side)

Therefore, triangle ABC is congruent to triangle ADC by ASA rule.

Now AB = DC and AD = BC (C.P.C.T)

Hence Proved.

THEOREM: If the opposite sides of a quadrilateral are equal, then it is a parallelogram.

A B

CDGiven: A quadrilateral ABCD in which AB=CD & AD=BC

To Prove: ABCD is a parallelogram.

Construction: Join A to C.

Proof: In triangle ABC and triangle ADC ,

AB = CD (given)

AD = BC (given)

AC = AC (common side)

Therefore triangle ABC is congruent to triangle ADC by SSS rule

Since the triangles of a quadrilateral are equal, therefore it is a parallelogram.

B

C

THEOREM: In a parallelogram opposite angles are equal.

A

DGiven: A parallelogram ABCD.

To prove: Angle A = Angle C & angle B=angle D

Proof: In the parallelogram ABCD,

Since AB is parallel to CD & AD is transversal

angles A+D=180 degrees (co-interior angles)-1

In the parallelogram ABCD,

Since BC is parallel to AD & AB is transversal

angles A+B=180 degrees (co-interior angles)-2

From 1 and 2,

angles A+D=angles A+B.

angle D= angle B.

Similarly we can prove angle A= angle C.

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

B

CD

A

Given: In a quadrilateral ABCD

angle A=angle C & angle B=angle D.

To prove: It is a parallelogram.

Proof: By angle sum property of a quadrilateral,

angles A+B+C+D=360 degrees

angles A+B+A+B=360 degrees (since, angle A=C and angle B=D)

2angle A+ 2angle B=360 degrees

2(A+B)=360 degrees

angles A+B= 180 degrees. (co-interior angles.)

Therefore, AD is parallel to BC

Similarly’ we can prove AB is parallel to CD.

This shows that ABCD is a parallelogram.

THEOREM: The diagonals of a parallelogram bisect each other.

B

CD

A

O

Given: A parallelogram ABCD

To prove: AO= OC & BO= OD.

Proof: AD is parallel to BC & BD is transversal.

angles CBD= ADB (alternate angles)

AB is parallel to CD & AC is transversal.

angles DAC= ACB (alternate angles)

Now, in triangles BOC and AOD,

CBD=ADB

DAC=ACB

BC=AD (opposite sides of a parallelogram)

Therefore, triangle BOC is congruent to triangle AOD by ASA rule.

Therefore, AO=OC & BO=OD [C.P.C.T]

This implies that diagonals of a parallelogram bisect each other.

THEOREM: If the diagonals of a quadrilateral bisect each other then it is a parallelogram.

B

CD

A

O

Given: In a quadrilateral ABCD,

AO = OC & BO = OD

To Prove: ABCD is a parallelogram.

Proof: In triangles AOD & BOC

AO = OC (given)

BO = OD (given)

angles AOD = BOC (vertically opposite angles)

Therefore, triangle BOC is congruent to triangle AOD by SAS rule

Therefore angle ADB = CBD & angle DAC = ACB (C.P.C.T)

Since alternate angles are equal, AD is parallel to BC.

Similarly, we can prove AB is parallel to CD.

This proves that ABCD is a parallelogram .

THEOREM: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

B

CD

A

Given: In a quadrilateral ABCD,

AB is parallel to CD AB = CD

To prove: ABCD is a parallelogram.

Construction: Join A to C.

Proof: In triangles ABC & ADC,

AB = CD ( given)

angle BAC = angle DCA (alternate angles.)

AC= AC ( common)

Therefore, triangle ABC is congruent to triangle ADC by SAS rule.

Therefore, angle ACB=DAC and AD=BC [C.P.C.T]

Since, AD is parallel to BC and AD=BC,ABCD is a parallelogram.

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

Given: A triangle ABC in which D and E are the mid- points of AB and Ac respectively.

To prove: DE is parallel to BC & DE=1/2BC

Proof: In triangles AED and CEF

AE = CE (given)

ED = EF (construction)

angle AED = angle CEF (vertically opposite angles)

Therefore, triangle AED is congruent to triangle CEF by SAS rule.

Thus, AD=CF [ C.P.C.T]

angle ADE = angle CFE [C.P.C.T]

C

A

B

DE

F

Now, AD= CF

Also, AD = BD

Therefore, CF = BD

Again angle ADE = angle CFE (alternate angles)

This implies that AD is parallel to FC

Since, BD is parallel to CF (since, AD is parallel to CF and BD=AD).

And, BD=CF

Therefore, BCFD is a parallelogram.

Hence, DF is parallel to BC and DF=BC (opposite sides of a parallelogram).

Since, DF=BC;

DE=1/2 BC

Since, DE=DF (given)

Therefore, DE is parallel to DF.

A

B C

DF

1

2

34

l

MA

E

Given: E is the mid- point of AB, line ‘l’ is passing through E and is parallel to BC and CM is parallel to BA.

To prove: AF=CF

Proof: Since, Cm is parallel to BA and EFD is parallel to BC, therefore BEDC is a parallelogram.

BE= CD( opposite sides of a parallelogram)

But, BE = AE, therefore AE=CD.

In triangles AEF & CDF: angle 1=2 (alt.angles)

angle 3=4 (alt.angles)

AE=CD (proved)

Therefore,triangle AEF is congruent to CDF(ASA)

AF=CF [C.P.C.T]. Hence, proved.

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