quadrilateral

Quadrilaterals

Quadrilaterals

Warm Up

Problem of the Day

Lesson Presentation

Warm UpIdentify the triangle based on the given measurements.

1. 3 ft, 4 ft, 6 ft

2. 46°, 90°, 44°

3. 7 in., 7 in., 10 in.

scalene

right

isosceles

Problem of the Day

One angle of a parallelogram measures 90°. What are the measures of the other angles?

all 90°

Learn to name and identify types of quadrilaterals.

Some quadrilaterals have properties that classify them as special quadrilaterals.

Quadrilateral just means "four sides" (quad means four, lateral means side).Any four-sided shape is a Quadrilateral.But the sides have to be straight, and it has to be 2-dimensional.

1.Parallelogram PropertiesQuadrilateral classification:

Quadrilaterals are classified according to the number of pairs of each parallel sides.

If a quadrilateral does not have any pair of parallel sides, it is called a TRAPEZIUM.

If a quadrilateral has only one pair of parallel lines, it is called TRAPEZOID.

If a quadrilateral has two pairs of parallel lines, it is called PARALLELOGRAM.A quadrilateral with opposite sides parallel and equal is a parallelogram . Properties:-• A diagonal of a parallelogram divides it into two congruent triangles.•In a parallelogram, opposite sides are equal.•In a parallelogram opposite angles are equal.•The diagonals of a parallelogram bisect each other.These properties have their converse also.

Special Parallelograms

If all the interior angles of a parallelogram are right angles, it is called a RECTANGLE.

If all the sides of a parallelogram are congruent to each other, it is called a RHOMBUS.

If a parallelogram is both a rectangle and a rhombus, it is called a SQUARE.

A Rhombus is a parallelogram with adjacentsides equal. The properties of rhombus are:-A rhombus has the including properties of A parallelogram.The diagonals of rhombus bisect each other at 90 degreeThe diagonals of rhombus bisect opposite angles

Properties of a Parallelograms

When GIVEN a parallelogram, the definition and theorems are stated as ...

A parallelogram is a quadrilateral with both pairs of opposite sides parallel.

If a quadrilateral is a parallelogram, the 2 pairs of opposite sides are congruent.

If a quadrilateral is a parallelogram, the 2 pairs of opposite angles are congruent.

If a quadrilateral is a parallelogram, the consecutive angles are supplementary.

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

If a quadrilateral is a parallelogram, the diagonals form two congruent triangles.

Mid-point theorem:-The line segment joining the mid point of

two sides of a triangle is always parallel to the third side and half of it.

Proof of mid point theorem

Given:-D and E are the mid points of the sides AB and AC .To prove:-DE is parallel to BC and DE is half of BC.construction:- Construct a line parallel to AB through C.proof:-in triangle ADE and triangle CFE AE=CE angle DAE= angle FCE (alternate angles ) angle AED= angle FEC (vertically opposite angles)Therefore triangle ADE is congruent to triangle CFE

Hence by CPCT AD= CF- - - - - - - - -1But AD = BD(GIVEN) so from (1), we get,BD = CFBD is parallel to CF Therefore BDFC is a parallelogramThat is:- DF is parallel to BC and DF= BC Since E is the mid point of DF DE= half of BC, and , DE is parallel to BC Hence proved .

Theorem : Sum of angles of a quadrilateral is 360

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.

A B

CD

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

Given: A parallelogram ABCD and its diagonal AC.

To prove: Triangle ABC is congruent to triangle ADC


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.A B

CD

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)


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.


Proof: In triangle ABC and triangle ADC ,

AB = CD (given)

AD = BC (given)


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.


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.

THANK YOU !!!

Submitted by: Alma BuisanSubmitted to: Ms. Charine Masilang

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quadrilateral