F.3 Maths/Ch.9/Quadrilaterals/CKH/p.1
1. If AB//CD, then x = y.
(alt.∠s, AB//CD)
A
2.
If x = y, then AB//CD.
(alt.∠s equal)
A
B
x
B
x
y
y
D
C
3. If AB//CD, then w + y = 180°.
(int.∠s, AB//CD)
4.
A
D
C
If w + y = 180°, then AB//CD.
(int.∠s, supp.)
A
B
B
w
w
y
C
C
D
5. If AB//CD, then z = y.
(corr.∠s, AB//CD)
6.
z
A
D
If z = y, then AB//CD.
(corr.∠s equal)
z
A
B
y
C
y
y
C
D
7. x + y = 180°
(adj.∠s on the st. line)
8.
D
x=y
(vert. opp.∠s)
C
A
x
x
y
y
A
B
9. c = a + b
(ext.∠ of  )
a
c
B
b
D
B
F.3 Maths/Ch.9/Quadrilaterals/CKH/p.2
Parallelogram
If the opposite sides of ABCD are parallel, then ABCD is a
parallelogram.
A
B
C
D
Properties of parallelogram
If ABCD is a parallelogram, then
AB = CD and AD = BC
(opp. sides of //gram)
A
B
C
D
If ABCD is a parallelogram, then
∠A = ∠C and ∠B = ∠D
(opp. ∠s of //gram)
A
B
C
D
If ABCD is a parallelogram and AC intersects BD at O,
then AO = CO and BO = DO.
(diags. of //gram)
A
B
O
Note: AO may not be equal to DO.
Tests for parallelogram
If AB = CD and AD = BC, then
ABCD is a parallelogram.
(opp. sides equal)
D
C
A
B
C
D
A
B
If ∠A = ∠C and ∠B = ∠D, then
ABCD is a parallelogram.
(opp. ∠s equal)
If AO = CO and BO = DO,
then ABCD is a parallelogram.
(diags. bisect each other)
B
O
D
A
If AB = CD and AB//CD, then
ABCD is a parallelogram.
(2 sides equal and //)
D
Test for rectangle
If ABCD is a parallelogram and ∠A = 90°,
then ABCD is a rectangle.
C
D
A
C
B
C
A
B
D
C
F.3 Maths/Ch.9/Quadrilaterals/CKH/p.3
Test for rhombus
1. If all sides are equal,
then ABCD is a rhombus.
A
B
D
C
2.
If ABCD is a parallelogram with two adjacent
sides equal, then ABCD is a rhombus.
A
B
D
Test for square
If all sides are equal and ∠A = 90°,
then ABCD is a square.
C
Properties of rectangle
Since a rectangle is also a parallelogram, it has
all properties of parallelogram.
Moreover, ∠A =∠B =∠C =∠D = 90°.
(property of rectangle)
And OA = OB = OC = OD.
(property of rectangle)
A
B
D
C
A
B
D
C
A
B
O
C
D
Properties of rhombus
Since a rhombus is also a parallelogram, it has
all properties of parallelogram.
A
B
D
Moreover, AC ⊥ BD. (property of rhombus)
And ∠OAD =∠OAB =∠OCB =∠OCD,
∠ODA =∠ODC =∠OBC =∠OBA
(property of rhombus)
Properties of square
Since a square is also a parallelogram, it has
all properties of parallelogram.
Moreover
AC ⊥ BD (property of square)
OA = OB = OC = OD (property of square)
∠ODC = 45° (property of square)
[the angle between any diagonal and a side is 45°]
C
A
O
D
B
C
A
B
O
D
45°
C
F.3 Maths/Ch.9/Quadrilaterals/CKH/p.4
Mid-point theorem
For triangle
If AB = BD, AC = AE, then
BC // DE, and
DE = 2BC
(mid-pt. theorem)
For trapezium
If AC = CE, BD = DF, then
CD//AB//EF, and
1
CD  ( AB  EF ) (mid-pt. theorem)
2
A
C
B
E
D
A
B
D
C
E
F
Equal ratios theorem
If AB // CD // EF, then
AC BD
AC CE


. (
)
CE DF
BD DF
(equal ratios theorem)
A
B
D
C
F
E
If AB // CD // EF // GH, then
AC : CE : EG = BD : DF : FH.
A
B
D
C
F
E
H
G
If BC // DE, then
AB AC

BD CE
A
C
B
D
Converse of equal ratios theorem
BD CE

In ∆ADE, if
, then
AB AC
BC // DE.
E
A
B
C
D
E
Intercept theorem
If AB // CD // EF and AC = CE, then
BD = DF.
A
B
C
D
F
E
A
If BC // DE and AB = BD, then
AC = CE.
B
Note : Since AB = BD and AC = CE,
DE = 2BC. ( mid-pt. theorem)
D
C
E
F.3 Maths/Ch.9/Quadrilaterals/CKH/p.5
Proof of mid-pt. theorem of triangle
In ∆ABC and ∆ADE,
∠BAC = ∠DAE
AB AC 1


AD AE 2
∴ ∆ABC ~ ∆ADE (ratio of 2 sides, inc∠)
BC 1

∴
(corr. sides, ~∆s)
DE 2
DE = 2BC
∠ABC = ∠ADE (corr. ∠s, ~∆s)
∴BC//DE (corr. ∠s equal)
A
C
B
E
D
Proofs of mid-pt. theorem of trapezium
To show that the mid-points of two non-parallel lines
is parallel to the parallel lines.
A
C
Proof:
Given: AC = CE
Draw AG which is parallel to BF and mark the mid-point H
of AG.
H
E
F
G
A
Then join CH which meets BF at D.
∵ AC = CE, AH = GH
∴ CH // EG (mid-point thm of triangle)
∴ CD // EF // AB
∵ AH // BD, AB // HD
∴ ABDH is a parallelogram.
∴ BD = AH (opp. sides, //gram)
B
C
B
H
E
D
F
G
Similarly, DF = HG
But AH = HG
∴ BD = DF
∴ D is a mid-pt of BF.
∴ the mid-points of two non-parallel lines is parallel to the parallel lines.
To show that the mid-points of two non-parallel lines
is half of the sum of the parallel lines.
A
Proof:
Let AB = b, CH = a.
EG = 2CH = 2a (proved)
AB = HD, HD = GF (opp. sides, // gram)
∴ AB = HD = GF = b
CD = CH + HD = a + b
EF = EG + CF = 2a + b
1
1
( AB  EF )  (b  2a  b)
2
2
=a+b
= CD
C
E
B
H
D
G
F
F.3 Maths/Ch.9/Quadrilaterals/CKH/p.6
Proof of equal ratios theorem
Draw AG and CH which are parallel to BF.
In ∆ACG and ∆CEH,
∠ACG = ∠CEH (corr.∠s, CD // EF)
∠CAG = ∠ECH (corr.∠s, AG // CH)
∴ ∆ACG ~ ∆CEH (AAA)
AC AG

∴
(corr. sides, ~∆s)
CE CH
A
B
D
C G
E
F
H
But AG = BD, CH = DF (opp. sides, // gram)
AC BD

∴
CE DF
A
AB AC

Can your prove that
if BC // DE ?
BD CE
B
C
D
Proof of converse of equal ratios theorem
Proof:
BD CE

∵
AB AC
BD
CE
1 
1
∴
AB
AC
BD  AB CE  AC

∴
AB
AC
AD AE

AB AC
∠BAC = ∠DAE ( com.∠)
∴ ∆ABC ~ ∆ADE ( ratio of 2 sides, inc∠)
∴∠ABC = ∠ADE ( corr. ∠s, ~∆s)
∴ BC // DE ( corr. ∠s equal)
E
A
B
D
C
E