Class IX Quadrilateral Chapter-8

1. PRESENTED BYPRESENTED BY
Shivam GoelShivam Goel
IX-EIX-E
Examination Roll No.-159212Examination Roll No.-159212

3. 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 360A B
C D
A B
C D

4. There are actually six types of quadrilaterals.
They are as follows:
 TRApEzIUm
 pARALLELogRAm
 REcTAngLE
 RhombUs
 sQUARE
 KITE

5. 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

6. PARALLELOGRA
M: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

7. 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

8. 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.

9. 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

10. 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.

11. 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 .

12.  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.

13. 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.

14. 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.

15. 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.
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.

16. THEOREM: In a parallelogram , opposite sides are
equal. D
A B
C
Given: 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)

17. 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.

18. 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.

19. 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.

20. 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.

21. 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 .

22. 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.

23. 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
D
E
F

24. 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.

25. A
B C
D
F
1
2
3
4
l
M
A
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.