1) The document provides information about various types of quadrilaterals including their properties and theorems related to rectangles and parallelograms. 2) Key quadrilaterals discussed include trapezoids, parallelograms, rectangles, squares, rhombus, and kites. Theorems are presented regarding the properties of diagonals and angles of rectangles. 3) Exercises with solutions are provided applying the theorems to problems involving finding angles of quadrilaterals, properties of rectangles and parallelograms, and relationships between sides and angles of figures.

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.