It is power point presentation for class 9 students , Chapter Quadrilaterals

1. SCHOOL NAME
Made By :- Master ?
Class :- ?
Roll no. :- ?
Subject :- Mathematics (ppt)

2. QUADRILATERALS
Quadrilateral :- Four sides polygon is called quadrilateral
having four sides
having four angels
having four vertices
Angel sum Property of :- Sum of all interior angles of is 360º
Given:-ABCD is a
To prove:-∠ A + ∠ D + ∠ B + ∠ C = 360°
Construnction:-Join B to D
Proof:- In Δ ADB
∠ A + ∠1 + ∠3 =180° - - - (1) [By angle sum property of Δ ]

3. In Δ BCD
∠C + ∠2 + ∠4 = 180° - - -(2)
Adding (1) & (2)
∠A + ∠c + (∠1 + ∠2) + (∠3 + ∠4) = 360°
∠A + ∠B + ∠C + ∠D = 360°
Types of Quadrilateral
 Trapezium :- A quad having one pair of Parallel side.
 Parallelogram :- A quad having two pair of || sides.
AD||BC
AB||CD , AD||BC
A B
CD
A B
CD

4.  Rectangle :- A IIgm having one angle of measure 90°
 Square :- A rectangle having adjacent sides are equal
 Rhombus :- A parallelogram having all sides equal
 A Kite :- A is a kite if it has two pairs of equal adjcent sides & unequal opposite sides
ABCD is IIgm∠B = 90°
B
A D
C
A
B C
D
A
A
B
B
C
C
D
D

5.  Theorem 1 : A diagonal of a IIgm divides it into two congruent triangles
 Given :- A IIgm ABCD
 To prove :- Δ ABC ≅ ΔCDA
 Proof :- Therefore , ABCD is a IIgm
AB II CD
∠1 = ∠2 (Alternate angle)
And AD II BC
∠3 = ∠4 (Alternate angle)
Now in Δ
Δ ABC & Δ ACD
∠2 = ∠1
∠4 = ∠3
AC = AC
Therefore , ΔABC ≅ ΔCDA
By ASA
A
B C
D

6.  Theorem 2 :- In a IIgm opposite sides are equal
 Given :- A IIgm ABCD
 To prove :- AB = CD & DA = BC
 Proof :- ABCD is a IIgm
AB II DC & AD II BC
And AD II BC
∠1 = ∠2 (Alternate angle)
And AB II DC
∠3 = ∠4 (Alternate angle)
Now in Δ
Δ ADC & ΔCBA
∠2 = ∠1
AC = AC
∠4 = ∠3
Δ ADC ≅ Δ CBA (By ASA)
AB = CB & DC = BA ( By CPCT)
A
B C
D

7.  Theorem 3 :- The Opposite angles of a IIgm are equal
 Given :- A IIgm ABCD
 To prove :- ∠A = ∠C & ∠B = ∠D
 Proof :- ABCD is IIgm
AB II DC & AD II BC
Now , AB II DC
Therefore , ∠A + ∠D = 180° - - - (1) [consecutive interior ∠ ]
And AD II BC
∠D + ∠C = 180° - - - (2)
From (1) & (2)
∠A + ∠D = ∠D + ∠C
∠A = ∠C
Similarly, ∠B = ∠D
Hence , ∠A = ∠c & ∠B = ∠D
A
B C
D

8.  Theorem 4 :- The diagonals of a IIgm bisect each other.
 Given :- A IIgm ABCD & diagonals AC & BD intersect O.
 To prove :- OA = OC , OB = OD
 Proof :- ABCD is a IIgm
AB II CD & AD II BC
∠1 = ∠2 [Alternate angle]
∠BAO = ∠DCO
& AB II DC
∠3 = ∠4
∠ABO = ∠CDO
Now in Δ
Δ AOB & ΔCOD
∠ BAO = ∠DCO
AB = CD
∠ABO = ∠CDO
By ASA , ΔAOB ≅ ΔCOD
OA = OC & OB = OD [By CPCT]
A
B C
D

9.  Theorem 5 :- In a IIgm , the bisectors of any two consecutive angles intersect at right angle.
 Given :- A IIgm ABCD , the bisectors of consecutive angle ∠A & ∠B intersect at P.
 To prove :- ∠APB = 90°
 Proof :- ABCD is a IIgm
AD II BC
∠A + ∠B = 180°
½ ∠A + ½ ∠B = 90°
∠1 + ∠2 = 90°
In Δ APB
∠1 = ∠APB = ∠2 = 180°
90° + ∠APB = 180°
∠APB = 90°
 Theorem 6 :- The diagonals of a IIgm bisect each other.
 Theorem 7 :- If the diagonals of a quadrilateral bisect each other , then it is a IIgm .
 Theorem 8 :- A quad. Is a IIgm if a pair of opp. Sides is equal & II .
A
B C
D

10. The Mid point theorem
Given :- ABC is a Triangle
To prove :- EF II BC
Proof :- In Δ ABC
EF = ½ BC
& ∠AEF = ∠ABC
So, EF II BC
 Theorem 9 :- The line segment joining the mid point of two sides of a triangle is II to the third side .
 Given :- A ΔABC in which D & F are the mid point of sides AB & AC respectively . DE is joined.
 To prove :- DE II BC & DE = ½ BC
 Construction:- Produce the line segments DE to F, Such that DE = EF .Join FC.
 Proof :- In Δs AED & CEF
AE = CE [ E is the mid-pt of AC]
∠AED = ∠CEF [ V.O.A]
DE = EF
A
B C
B
A
C
D
E
E
F
F

11. Δ AED ≅ Δ CEF [ by SAS ]
AD = CF , ∠ADE = ∠CFE [CPCT]
Now , D is the mid-pt of AB
AD = DB
DB = CF - - - - (1)
Now , DF intersect AD & EC at D and F respectively such that
∠ADE = ∠CFE [ Alternate ∠ ]
AD II FC
DB II CF - - - -(2)
From (1) & (2)
therfore , DBCF is A IIgm
DF II BC & DF = BC
D ,E , F are collinear & DE =EF
DE II BC & DE =1∠2 BC

12.  Theorem 10 :-The line drawn through the mid-pt of one side of a triangle , parallel to another side bisects the
third side.
 Given :- Δ ABC in which D is the mid-pt of AB & DE II BC
 To prove :- E is the mid-pt of AC
 Proof :- We have to prove that E is the mid-pt of AC . If possible , let E be not the mid-pt of AC . Let E prime be
the mid-pt AC. Join DE prime .
Now , in Δ ABC , D is the mid-pt of AB [ Given ]
And E prime is the mid-pt of AC .
DE’ II BC - - - (1)
DE II BC - - - (2)
From (1) & (2)
We find that two intersecting lines DE & DE’ are both II to line DC
This is a contradiction to the II line axiom .
So, our supposition is wrong . Hence , E is the mid-pt of AC .

13. EXERCISE
[1] The angles of quad. are in the ratio 3:5:9:13 . Find all the angles of the quad.
Sol: let the angles be (3x),(5x),(9x),(13x)
3x + 5x + 9x + 13x =360 °
30x=360 °
x =360/30
x=12 °
therefore, the angeles are (3*12),(5*12),(9*12) & (13*12)
i.e. 36 ° , 60 ° , 108 ° & 156 °
[2] If the diagonals of a IIgm are equal , then show that it is a rectangle .
Given :-A IIgm ABCD in which AC =BD
To prove :-ABCD is a rectangle
Proof :-In Δs ABC &DCB A B
CD

14. AB= DC
BC=BC
AC=DB
Δ ABC ≅ Δ DCB [by SSS]
∠ABC = ∠DCB [by CPCT]
[3] Diagonals AC of a llgm ABCD bisects ∠A.
i. It bisects ∠C also,
ii. ABCD is a rhombus
Sol: Given :-A IIgm ABCD in which diagonals AC bisects ∠A
To prove :- that AC bisects ∠C
Proof :- ABCD is a IIgm
AB II DC
Now , AB II DC & AC intersects them
A B
CD

15. ∠1= ∠3 - - -(1) [alternate interior angle]
AD II BC & AC intersects them
∠2=∠4 - - -(2)
Ac is bisector of ∠a
∠1=∠2 - - -(3)
From (1) ,(2)&(3)
∠3=∠4
Hence, AC is bisector ∠C
To prove :-ABCD is a rhombus
Proof :- (1),(2)&(3) give ∠1=∠2=∠3=∠4
Now in triangle ABC
∠1=∠4
AB=BC
In Δ ADC
AD=DC
ABCD is IIgm
AB=CD,AD=BC

16. combining these
AB=BC=CD=DA
Hence, ABCD is a rhombus.
[4] In a IIgm ABCD , two points P & Q are taken on diagonal BD such that DP = BQ. show that :
i. Δ APD ≅ Δ CQB
ii. AP = CQ
iii. Δ AQB ≅ ΔCPD
iv. AQ = CP
v. APCQ is a IIgm
Sol: Given :- ABCD is a IIgm . P & Q are points on the diagonals BD such that DP = BQ.
To prove :-Δ APD ≅ Δ CQB
AP = CQ
Δ AQB ≅ ΔCPD
AQ = CP
APCQ is a IIgm
Proof :- (1) In Δs APD & CQB
AD = CB [ opp. Sides of a IIgm ABCD]
A
B C
D
Q
P

17. AP = CQ
DP = BQ
ΔAPD ≅ ΔCQB [By SSS]
(2) AP = CQ [opp. Sides of a IIgm APCQ]
(3) In Δs AQB & CPD
AB = CD
AQ = CP
BQ = DP
Δ AQB ≅ ΔCPD [By SSS]
(4) AQ = CP [ Opp. Sides of a llgm APCQ]
(5) AQ = CP {CPCT}
AP = CQ {CPCT}
Hence, APCQ is a llgm
[5] ABCD is trapezium in which AB ll CD & AD = BC
i. ∠A = ∠B
ii. ∠C = ∠D
iii. ΔABC ≅ Δ BAD
iv. Diagonals AC = diagonals BD
A B
CD
E

18. Sol: Given :- ABCD is a trapezium in which AB ll CD & AD = BC.
To prove :- ∠A = ∠B
∠C = ∠D
ΔABC ≅ Δ BAD
Diagonals AC = diagonals BD
Construction :- Produce AB and draw a line CE ll AD
Proof :- (1) Since AD ll CE & transversal AE cuts them at A & E respectively
therefore ∠A + ∠E = 180 ° - - - (1)
AB ll CD & AD ll CE , therefore AECD is a llgm
AD = CE
BC = CE
In Δ BCE
BC = CE
∠CBE = CEB
180 - ∠B = ∠E
180- ∠ E = ∠B - - -(2)

19. From (1) & (2)
∠A = ∠B
(2) ∠A = ∠B
∠BAD = ∠ABD
180 - ∠BAD = 180 - ∠ABD
∠ADB = ∠BCD
∠D = ∠C i.e. ∠C = ∠D
(3) In Δs ABC & BAD
BC = AD
AB = BA
∠A = ∠B
ΔABC ≅ ΔBAD [By SAS]
(4) Since ΔABC ≅ ΔBAD
AC = BD [CPCT]

20. [6] ABCD is a trapezium in which AB ll DC , BD is a diagonal and E is the mid-pt of
AD . A line is drawn though E parallel to AB intersecting BC at F . Show that F is the
mid-pt of BC.
Sol: Given :-In trapezium ABCD , AB ll DC , E is the mid-pt of AD,EF ll AB.
To prove :- F is the mid-pt of BC
Proof :- In Δ DAB, E is the mid-pt of Ad
EG ll AB
therefore, by converse of mid-pt theorem G is the mid-pt of DB.
In ΔBCD , G is the mid-pt of
GF ll DC
therefore, by converse of mid-pt theorem
F is the mid-pt of BC.