Booklet-2_Lines-Angles-Triangles-Quads-11-March-2021.pdf

Maths study guide

1. Lines, Angles, Δ
s
& Quadrilaterals
(Grade 8 to 10 Revision)
by The TAS Maths Team
TAS FET EUCLIDEAN GEOMETRY COURSE
BOOKLET 2
Lines, Angles, Δ
s
&
Quadrilaterals
Geometry FET
Course Booklets Set
WWW.THEANSWER.CO.ZATAS

2. CONTENTS of Booklet 2 Lines & Angles  Vocabulary and Facts Triangles

 Vocabulary, Facts and Proofs  Area Quadrilaterals

 Properties  Definitions  Theorems  Area Grade 10: Midpoint Theorem  Statement and Converse  Riders
EXERCISES & FULL SOLUTIONS on all 4 sections
TAS

3. Copyright © The Answer Series

1




The Language (Vocabulary)

C PARALLEL LINES C PERPENDICULAR LINES AB || CD

AB ⊥ CD





C SINGLE ANGLES







C PAIRS OF ANGLES
•
Complementary
ø
s
add up to 90º
e.g. 40º and 50º;
x
and 90º –
x



•
Supplementary
ø
s

add up to 180º
e.g. 135º and 45º;
x
and 180º –
x




•
Adjacent
ø
s
have a common vertex
and a common arm and lie on
opposite sides of the common arm.


e.g. A pair of adjacent A pair of adjacent

complementary ø
s
: supplementary ø
s
:




WHEN 2 LINES INTERSECT, WE HAVE . . .

(
adjacent
ø
s
&
(
vertically opposite

ø
s
,

1 and 4
ˆ ˆ
;
1 and 3
ˆˆ


4 and 3
ˆˆ
, etc. or
2 and 4
ˆ ˆ





WHEN 2 LINES ARE CUT BY A TRANSVERSAL . . .


2 'families'
of 4 angles are formed


C PAIRS OF ANGLES
, one from each family:
• corresponding ø
s
:
1
&
5
;
2
&
6
;
3
&
7
;
4
&
8

• alternate ø
s
:
3
&
5
and
4
&
6

• co-interior ø
s
:
4
&
5
and
3
&
6


( Corresponding
means their positions correspond.

( Alternate means:


ø
s

lie on opposite sides of the transversal;
whereas, 'co-' means: 'on the same side of the transversal'.

e.g.
corresponding alternate co-interior


angles

angles angles




The ANGLE is the
amount of rotation
about the vertex.
acute
angles
obtuse
angles
reflex
angles
0º 90º 180º 270º 360º
a right angle a straight angle a revolution
A
B
C
D
A B
CD
arms vertex
NO
T
E: The
p
lural o
f
vertex is vertice
s
!
40º 50º
x

90º – x
135º
45º
LINES, ANGLES & TRIANGLES



A
C
B
D
D
A
C B
90º
40º
140º
50º
40º
!
Know the meanings
of all the WORDS.
1 2 3 4
5 6 7 8
1
2
3
4
x

180º – x
7
1
2
4
3
5 6
8
common arm
common
vertex
LINES: 2 SITUATIONS


the
transversal
1 2
4 3
5
6
8 7 TAS

4. 2
Copyright © The Answer Series



Classification according to . . .


SIDES

ANGLES


Scalene Δ

Acute ø
d
Δ


(all 3 sides different in length)

(all 3 ø
s
are acute)

Isosceles Δ

Right ø
d
Δ

(2 sides equal in length)

(one ø = 90º)

Equilateral Δ

Obtuse ø
d
Δ

(all 3 sides equal in length)

(one ø is obtuse)



In an isosceles triangle:




We can classify Δ
s
according to sides and ø
s
simultaneously:

Examples



This is a/an . . . . -angled Δ This is a/an . . . . -angled Δ This is a/an . . . . -angled Δ




















The Facts


LINES





INTERSECTING LINES


When two lines intersect, any pair of adjacent angles is
supplementary.

e.g. 4
ˆ
+ 1
ˆ
= 180º & 4
ˆ
+
3
ˆ
= 180º



Vertically opposite angles are equal .
e.g. 1
ˆ
=
3
ˆ
and 2
ˆ
= 4
ˆ










PARALLEL LINES

When ANY 2 lines are cut by a transversal, 2 'families' of four ø
s
are formed,
and there are:
corresponding ø
s
;
alternate ø
s
;
& co-interior ø
s
. . . whether the lines are parallel or not !
If 2 PARALLEL lines are cut by a transversal,
corresponding angles are equal ;
alternate angles are equal ; and
co-interior angles are supplementary.

1 2
3 4
5 6
7 8
1 2
3 4
2 intersecting lines
form 4 angles
the vertical angle
the base angles
BASE




Fact 2

Fact 1

Fact 3

The sizes of the
other angles?



Special case:
Perpendicular lines


TRIANGLES



Answers 

An isosceles right-ø
d
Δ


An isosceles acute-ø
d
Δ


A scalene obtuse-ø
d
Δ
Interior and Exterior angles


Interior
angles: &
Exterior
angles:


x
y
z
a
c
b
d
f

e
o
r

A
n exterior angle
is formed between
a side of the triangle
and the produced
(extension) of
an adjacent side of
the triangle.

x

x
is not an exterior ø
. . . the side is not 'produced'.
1 2
3 4
5 6
7 8 TAS

5. Copyright © The Answer Series

3

The CONVERSE STATEMENT of FACT 3 says
If a pair of corresponding angles is equal

OR

If a pair of alternate angles is equal

OR

If a pair of co-interior angles is supplementary
then:
line AB is PARALLEL to line CD
. . . whether it looks like it or not!


FACT 1

says:
If ABC is a straight line, then 1
ˆ
+ 2
ˆ
= 180º.








Example
If, in the sketch alongside,
x

= 40º and y = 130º,
is PQR a straight line?







Now, refer to FACT 3 on the previous page . . .


















The Facts, continued


TRIANGLES




FACT

1: The Sum of the interior angles of a triangle . . .


The sum of the (interior) angles
of a triangle is 180º.


FACT 2: The Exterior angle of a triangle . . .


The

exterior angle of a triangle equals
the sum of the interior opposite angles.


FACT 3: An Isosceles triangle . . .


In

an isosceles triangle,
the base angles are equal.






Example
If in isosceles ΔABC,
the vertical angle (A)
ˆ
is a right ø,
determine the size of the base ø
s
.



If AB = AC,
then: 1
ˆ
= 2
ˆ

y
x
P
Q
R
CONVERSE STATEMENTS

A
ˆ
+ B
ˆ
+
C
ˆ
= 180º
A

B C
1 2
A

B C
1
ˆ
= 2
ˆ
+
3
ˆ
2
3 1
The CONVERSE STATEMENT of FACT 1 says
If 1
ˆ
+ 2
ˆ
= 180º, then: ABC is a straight line.

A

B
C
D
So, conversely:
If 1
ˆ
= 2
ˆ
,
then: AB = AC
The CONVERSE states:

If 2 angles of a triangle are equal,
then: the sides opposite them are equal.
x

A

B C
x
1
2
A

B
C

Answer:
x
+ y = 40º + 130º = 170º ≠ 180º
â No, PQR is not a straight line . . .

Whether it looks
like it or not!
Answer
2
x
= 90º
∴
x
= 45º TAS

6. 4
Copyright © The Answer Series


The CONVERSE states:
If the square on one side
of a triangle equals the
sum of the squares on the
other two sides
(in area)
,
then the triangle is
right-angled.

FACT 4: An Equilateral triangle . . .


The

angles of an equilateral triangle
all equal 60º.


FACT 5: THE THEOREM OF PYTHAGORAS

THE

THEOREM OF PYTHAGORAS states:
The square on the hypotenuse of a right-angled triangle
equals the sum of the squares on the other two sides
(in area)
.











Note:
• Only one angle can be 90º (a right angle)
• The side opposite the right-angle is called the hypotenuse.

FACT 6: The Area of a triangle

The

area of a triangle =
base height
2
%




FACT 7: Similar triangles
Definition of Similarity Two figures are SIMILAR if:


they are equiangular, and


their corresponding sides are in proportion.

e.g.



In the case of TRIANGLES (only):
If triangles are equiangular,
then: their corresponding sides are in proportion
and therefore: they are similar.

i.e. If A
ˆ
= D
ˆ
, B
ˆ
= E
ˆ
and
C
ˆ
=
F,
ˆ

then:
AB
DE
=
BC
EF
=
AC
DF
,
and therefore: ΔABC |||
ΔDEF.















60º
60º 60º
D
E F
A

B C
3
x
= 180º
∴
x
= 60º
The CONVERSE states:
If the sides of two triangles are in proportion ,
then: these triangles will also be equiangular
and therefore: the triangles are similar.
i.e. If
AB
DE
=
BC
EF
=
AC
DF
,
then: A
ˆ
= D
ˆ
, B
ˆ
= E
ˆ
and
C
ˆ
= F
ˆ
,
and therefore: ΔABC
|||
ΔDEF
If
ˆ
C
= 90º,
then:
c
2
= a
2
+ b
2


Conversely:
If c
2
= a
2
+ b
2
,
then:

ˆ
C
= 90º
A

C
B a
b
c
h
b
The symbol:
|||

trian
g
les

quadrilaterals

penta
g
ons

. . .

2 conditions
TAS

7. Copyright © The Answer Series

5

FACT 8: Congruent triangles


Two

triangles are congruent if they have
• 3 sides the same length . . .
SSS

• 2 sides & an included angle equal . . .
SøS

• a right angle, the hypotenuse & a side equal . . .
RHS
or
SS90º

• 2 angles and any side equal . . .
øøS






If we can prove that ΔABC ΔDEF, then we can conclude that
all the sides and angles are equal.









Congruency can be understood best by constructing triangles,
even casually imagining the construction!







The possibilities are:


SSS
– only one size (& shape) of triangle could
be constructed.


SSø
– the case where the angle is
NOT INCLUDED


between the sides.

This is the ambiguous case because
there are 2 possible
∆
s

which we could draw

RHS
–

the angle is not included,
but it is a right angle,
so only 1 option is possible:



SøS
– the angle is included.
Only 1 option is possible.



øøS
–

given 2 angles, we actually have all 3 angles since the sum of
the angles must be 180º.

The side restricts the size of the triangle.

So, only 1 option is possible.

However, when comparing triangles,
the given equal sides must correspond
in relation to the angles.


øøø –
A side is
required!

Any number of possible options.
Congruent Δ
s
have the same SHAPE, and the same SIZE.
Similar Δ
s
have the same SHAPE, but not necessarily the same SIZE.






o
r
o
r

obtuse
ø
d
∆
acute
ø
d
∆

A

B C
D
F E
o
r

The symbol:

Conditions of Congruency

A triangle has 6 parts, 3 sides and 3 angles, which can be measured.
However, we use 3 measurements at a time to construct a triangle.
TAS

8. 6
Copyright © The Answer Series



( If 2 triangles are
equal in every respect
(i.e. all 3 sides and
all 3 angles), we say they are
congruent
.

We write: ABC

PQR.

( Enlarging or reducing a triangle, as on a copier, will produce a triangle

similar
to the original one.
i.e. It will have the
same shape
(all angles will be the same size as before)
but the respective sides will not be the same length.
The sides will, however, be proportional.

We write: ABC ||| PQR.






















































Comparing Triangles – Congruence vs Similarity

Proofs of Triangle Facts 1 and 2


Triangle Fact 1
Why is the sum of the interior angles of a triangle 180º?

See ΔABC:




Can we prove that
B
ˆ
+
A
ˆ
+
C
ˆ
= 180º ?
A

B C

Draw line DAE through A, parallel to BC.














We know:
1
ˆ
+
2
ˆ
+
3
ˆ
= 180º . . .
adjacent ø
s
on a straight line, DAE

But
ˆ
1
= alternate
ˆ
B
&
ˆ
3
= alternate
ˆ
C
. . .
DAE || BC

â
ˆ
B
+
ˆ
2
+
ˆ
C
= 180º
D
A

B C
E
1
2
3
Triangle Fact 2

Why is the exterior angle of a triangle equal to
the sum of the two interior opposite angles?







xˆ
+
ˆ
y
= 180º
. . . see LINES fact 1 (Page 2)

but
xˆ
+
ˆˆ
(A + B)
= 180º
. . . see TRIANGLES fact 1 (Page 3)

â
ˆ
y
=
ˆˆ
A + B

Logical !
A

B
C
D
x

y TAS

9. Copyright © The Answer Series

7

QUADRILATERALS


The Facts

C Properties of Quadrilaterals Recall all the quadrilaterals ...
(kite, trapezium, parallelogram, rectangle, rhombus, square)
.
What properties do they have?

C Equal sides?











C Parallel sides?







C Angles? Which are equal? Which are supplementary? Which are right angles?
C Diagonals?











Investigating quadrilaterals, using diagonals:
fig. 1: Use a diagonal to determine the sum of
the interior angles of a quadrilateral.

fig. 2: Use a diagonal to find the area of
a trapezium.

fig. 3 - 6: Which of these quadrilaterals have their areas bisected by
the diagonal?

fig. 3 - 6: Draw in the second diagonal. For each figure, establish whether the
diagonals are:

C equal C bisect each other
C intersect at right angles C bisect the angles of the quadrilateral

fig. 6: Find the area of a kite in terms of its diagonals.
Could this formula apply to a rhombus? A square?
C Defining Quadrilaterals C A trapezium



C A parallelogram We have observed the
properties
of a parallelogram:
C both pairs of opposite sides parallel
C both pairs of opposite sides equal
C both pairs of opposite angles equal
C diagonals which bisect one another.



Observe the progression of quadrilaterals below as we discuss further definitions:







Which property does a parallelogram need to become a rectangle?
Which property does a parallelogram need to become a rhombus?
Which property does a rectangle need to become a square?
Which property does a rhombus need to become a square?
square
parallelogram
rhombus kite
rectangle
A diagonal of a
quadrilateral is
a line joining
opposite vertices.
rectangle
rhombus
trapezium
parallelogram
square
'any'
quadrilateral
How would you find the sum
of the interior angles of
a pentagon? A hexagon?
We will, however,
define
the
parallelogram in terms of
its
p
arallel lines.
'any' quadrilateral
kite
rhombus
square
Definition: A parallelogram is a quadrilateral with

TWO PAIRS OF OPPOSITE SIDES
parallel.
o
r
1
3
5
6a
6b
2
4a
4b
trapezium
parallelogram
square
'any'
quadrilateral
rectanglerhombus
Definition: A trapezium is a quadrilateral with

ONE PAIR OF OPPOSITE SIDES
parallel. TAS

10. Copyright © The Answer Series

8

C A rectangle




C A rhombus





C A square










All these definitions are extremely important to know when you're doing sums.

e.g. If asked to prove that a particular quadrilateral is a rectangle, and you already
know it is a parallelogram, all you need to show is that one angle is a right angle.

Observe the following progression of quadrilaterals before the next definitions:









C A kite

Is a rhombus a kite?

A
rhombus
is a kite with .....................................................................................

A
square
is a kite with ........................................................................................




See the summary of quadrilaterals on the next page:

'Pathways of definitions, areas and properties'


Observing the
progression
of quadrilaterals – along routes 1 & 2 – is essential for
understanding
definitions, properties (especially diagonals) and area formulae.

None of these facts and proofs should need to be memorised.


C Proving conjectures







e.g. 1 Conjecture: The diagonals of a rhombus bisect the angles of the rhombus.

Proof:

ˆ
x
1

=
ˆx
2

. . . ø
s
opp = sides

=
ˆ
x
3
. . .
alt ø
s
; || lines

=
ˆ
x
4
. . .
ø
s
opp = sides



e.g. 2 Conjecture: The diagonals of a rhombus intersect at right angles.
(The proof is on the
SUMMARY
on the next page)
Note:
The further back you go on the route,
the more the definition re
q
uir
e
s!
Note: The curriculum says:
Define the quadrilaterals.
Investigate and make conjectures about the properties of the
sides, angles, diagonals and areas of these quadrilaterals.

Prove these conjectures.
x
2
x
1
x
3
x
4
'any' quadrilateral
kite
rhombus
square
Definition: A square is a rhombus with . . . . . . . . . . . . . .
A square is a rectangle with . . . . . . . . . . . . . .
A square is a parallelogram with . . . . . . . . . . . . . .
A square is a quadrilateral with . . . . . . . . . . . . . .
Definition: A rectangle is a parallelogram with 1 right angle
OR: A rectangle is a quadrilateral with . .?. . right angles?
Definition: A rhombus is a parallelogram with a pair
of adjacent sides equal

OR: A rhombus is a quadrilateral with . .?. . equal sides?
Definition: A kite is a quadrilateral with 2 pairs of adjacent sides equal.
Definitions tell us what the minimum
(
least
)
is that one needs! TAS

11. Copyright © The Answer Series

9

QUADRILATERALS
- pathways of definitions, areas and properties - A Summary








































'Any' Quadrilateral
A Trapezium
A Parallelogram
The Square
A Rhombus
A Kite
DEFINITION : Quadrilateral with 1 pair of opposite sides ||
DEFINITION : Quadrilateral with 2 pairs opposite sides ||
DEFINITION : A ||
m
with one pair of
adjacent sides equal
the 'ultimate' quadrilateral !
Sum of the ø
s
of
an
y
q
uadrilatera
l
= 360º
Area = s
2

Area = Δ 1 + Δ 2
=
1
2
ah +
1
2
bh
=
1
2
(a + b) . h

'Half the sum of the || sides
% the dis
t
ance between them.
'
Properties :
It's all been said 'before' !

Since a square is a rectangle,
a rhombus, a parallelogram,
a kite, . . . ALL the properties
o
f
these
q
uadrilaterals a
pp
l
y
.
Area
=
1
2
product of diagonals (as for a kite)
or
= base % height (as for a parallelogram)

Given diagonals a and b
Area = 2 Δ
s
= 2

1a
b
22
⎛⎞
⎜⎟
⎝⎠
.
=
ab
2

'Half the product of the diagonals'

THE DIAGONALS
bisect one another
PERPENDICULARLY

bisect the angles of the rhombus
bisect the area o
f
the rhombus
THE DIAGONALS
cut perpendicularly
the
LONG DIAGONAL
bisects:
the short diagonal, the opposite angles and the area of the kite
Quadrilaterals
play a
prominent role
in both Euclidian
and Analytical
Geometry right
through to
Grade 12
2!
a
b
c
d
e
f

h
b
a
1
2
b
a
2

a
2

Sum of the interior angles
= (a + b + c) + (d + e + f)
= 2 % 180º . . . (2 Δ
s
)
= 360º
Properties :
2 pairs opposite sides equal
2 pairs opposite angles equal
&
DIAGONALS BISECT ONE ANOTHER

Area = base % height






||
m
ABCD = ABCQ + ΔQCD
rect. PBCQ = ABCQ + ΔPBA
where ΔQCD ≡ ΔPBA . . .
RHS
â ||
m
ABCD = rect. PBCQ (in area)
= BC % QC

P
A Q
D
BC
All you need to know!
DEFINITION : Quadrilateral with 2 pairs of
adjacent sides equal
A Rectangle
DEFINITION : A ||
m
with one right ø
Area =
ℓ
% b
DIAGONALS are E
Q
UAL
angles diagonals sides
Note :
y
y
y
y
x

x
x

x

2
x
+ 2y = 180º . . .
ø
s
of , or

C
x
+ y = 90º
co-int. ø
s
suppl
.
The arrows indicate
various ‘pathways’
from ‘any’
quadrilateral to the
square (the ‘ultimate
quadrilateral’). These
pathways, which combine
logic and fact, are
essential to use when
proving specific types
of quadrilaterals.

See how the properties
accumulate as we
move from left to right.
i.e. the first quad has
no special properties
and each successive
quadrilateral has all
p
recedin
g
p
ro
p
erties. TAS

12. Copyright © The Answer Series
1 0

o
AN ASSIGNMENT
o


TASK A:

Theorems 1 → 3

Prove each of these properties yourself,
STARTING WITH THE DEFINITION

as the 'given'.

TASK B:
Theorems 4 → 7

Prove these four converse theorems,

WORKING TOWARDS THE DEFINITION ,
i.e. you need to prove, given any one of these situations, that the quadrilateral
would have 2 pairs of opposite sides parallel, i.e. that,
by definition
, the
quadrilateral is a parallelogram.
Hint
Use your FACTS on II lines
and congruent triangles.
C
Theorems and Proofs

The following section deals with the properties of a parallelogram. We firstly prove
all the properties. Secondly, we prove that a quadrilateral with any of these
properties has to be a parallelogram.
Geometry is an exercise in LOGIC. Initially, we observe, we measure, we
record . . . But, finally . . . We decide on how to define something and
then we prove various properties logically, using the definition.








Beyond the DEFINITION of a parallelogram, we noticed other facts/properties
regarding the lines, angles and diagonals of a parallelogram. The statement and
proofs of these properties make up our first three THEOREMS!



The PROPERTIES of a parallelogram


Theorem 1: The opposite angles of a parallelogram are equal.
Theorem 2: The opposite sides of a parallelogram are equal.
Theorem 3: The diagonals of a parallelogram bisect one another.


The CONVERSE theorems
Given a property, prove the quadrilateral is a parallelogram,
i.e. prove both pairs of opposite sides are parallel.

There are four converse statements, each claiming that IF a quadrilateral
has a particular property, it must be a parallelogram.


Theorem 4: If a QUADRILATERAL has 2 pairs of opposite angles equal,
then the quadrilateral is a parallelogram.

Theorem 5: If a QUADRILATERAL has 2 pairs of opposite sides equal,
then the quadrilateral is a parallelogram.

Theorem 6: If a QUADRILATERAL has 1 pair of opposite sides equal and
parallel, then the quadrilateral is a parallelogram.

Theorem 7: If a QUADRILATERAL has diagonals which bisect one another,
then the quadrilateral is a parallelogram.



































In these cases, we work
towards the definition
!
THE DEFINITION OF A PARALLELOGRAM
A parallelogram is a quadrilateral with
2 PAIRS OF OPPOSITE SIDES PARALLEL.
All the properties are to be deduced
from the definition!
TAS

13. 1 1

Copyright © The Answer Series


The Theorem Proofs

THE PROOFS OF THE PROPERTIES












Theorem 1: The opposite angles of a ||
m
are equal.

Given: ||
m
ABCD
i.e. AB || DC and AD || BC

RTP:
ˆ
A
=
ˆ
C
and
ˆ
B
=
ˆ
D


Proof:
ˆ
A
+
ˆ
B
= 180º . . .
co-interior ø
s
; AD || BC

But,
ˆ
A
+
ˆ
D
= 180º . . .
co-interior ø
s
; AB || DC

â
ˆ
B
=
ˆ
D

Similarly,
ˆ
A
=
ˆ
C


Theorem 2: The opposite sides of a ||
m
are equal.

Given: ||
m
ABCD
i.e. AB || DC and AD || BC

RTP: AB = CD and AD = BC

Construction: Draw diagonal AC . . .

Proof: In Δ
s
ABC and ADC
1)
ˆ
1
=
ˆ
2
. . .
alternate ø
s
; AB || DC

2)
ˆ
3
=
ˆ
4
. . .
alternate ø
s
; AD || BC

3) AC is common
â ΔABC ≡ ΔCDA . . .
øøS

â AB = CD and AD = BC


Theorem 3: The diagonals of a parallelogram bisect
one another.
Given: ||
m
ABCD with diagonals AC and BD intersecting at O.

RTP: AO = OC and BO = OD

Proof: In Δ
s
AOB and DOC
1)
ˆˆ
1 = 2
. . .
alt ø
s
; AB || DC

2)
ˆˆ
3 = 4
. . .
vert opp ø
s

3) AB = DC . . .
opposite sides of ||
m
– see theorem 2 above

â ΔAOB ≡ ΔCOD . . . øøS
â AO = OC and BO = OD










THE CONVERSE PROOFS Theorem 4: If a QUADRILATERAL has 2 pairs of opposite
angles equal, then the quadrilateral is a ||
m


Given: Quadrilateral ABCD with
ˆ
A
=
ˆ
C
and
ˆ
B
=
ˆ
D


RTP: ABCD is a parallelogram,
i.e. AB || DC and AD || BC

Proof: Let
ˆ
A
=
ˆ
C
=
x
and
ˆ
D
=
ˆ
B
= y
then
ˆ
A
+
ˆ
B
+
ˆ
C
+
ˆ
D
= 360º . . .
sum of the ø
s
of a quadrilateral

â 2
x
+ 2y = 360º
÷ 2) â x
+ y = 180º

i.e.
ˆ
A
+
ˆ
D
= 180º and
ˆ
A
+
ˆ
B
= 180º
â AB || DC and AD || BC . . .
co-interior ø
s
are supplementary

â ABCD is a parallelogram . . .
both pairs of opposite sides ||

DON'T EVER MEMORISE THEOREM PROOFS!
Develop the proofs/logic for yourself
be
f
or
e
checking against the methods shown below.
Note: We used the result in theorem 2 in the proof of theorem 3 –
but, we could've started from the beginning,
i.e. from the definition of a parallelogram.
We would just have needed to prove an extra pair of Δ
s

congruent (as in theorem 2) .
Theorems:
Definition

Property
Converse

theorems:
Property

Definition

Make sense of
THE LOGIC!

RTP: Required to prove
We could, of course, also have proved the first theorem this way!
It doesn’t matter which
diagonal you draw

1
A
B
C D
4
>
>
2
3
A
B
C D
C D
4
1
2
3
>
>
A
B
A

>
>
B
D
x

y
C
x

y
TAS

14. Copyright © The Answer Series
1 2

C Theorem 5: If a QUADRILATERAL has 2 pairs of opposite
sides equal, then the quadrilateral is a ||
m


Given: Quadrilateral ABCD with AB = CD
and AD = BC

RTP: ABCD is a parallelogram,
i.e. AB || DC and AD || BC

Construction: Draw diagonal AC . . .
it doesn’t matter which diag. you draw


Proof: In Δ
s
ACD and CAB
1) AC is common
2) AD = BC . . .
given

3) CD = AB . . .
given

â ΔACD ≡ ΔCAB . . .
SSS

â
ˆ
1
=
ˆ
2
and
ˆ
3
=
ˆ
4

â AB || DC and AD || BC . . .
alternate ø
s
are equal

â ABCD is a parallelogram . . .
both pairs of opposite sides ||


C Theorem 6: If a QUADRILATERAL has 1 pair of opposite
sides equal and ||, then the quadrilateral is a ||
m


Given: Quadrilateral ABCD with AB = and || DC

RTP: ABCD is a parallelogram,
i.e. AB || DC and AD || BC

Construction: Draw diagonal AC . . .

Proof: In Δ
s
ABC and CDA
1) AB = DC . . .
given

2)
ˆ
1
=
ˆ
2
. . .
alternate ø
s
; AB || DC

3) AC is common
â ΔABC ≡ ΔCDA . . .
SøS

â
ˆ
3
=
ˆ
4

â AD || BC . . .
alternate ø
s
equal


But AB || DC . . .
given

â ABCD is a parallelogram . . .
both pairs of opposite sides ||

C Theorem 7: If a QUADRILATERAL has diagonals which
bisect one another, then the quadrilaterals
is a ||
m
.

Given: Quadrilateral ABCD with diagonals AC and BD intersecting at O and
AO = OC and BO = OD.

RTP: ABCD is a parallelogram,
i.e. AB || DC and AD || BC

Proof: In Δ
s
AOB and COD
1) AO = OC . . .
given

2)
ˆˆ
1 = 2
. . .
vert opp ø
s


3) BO = O . . .
given

â ΔAOB ≡ ΔCOD . . .
SøS

â
ˆˆ
BAD = OCD

â AB || DC . . .
alternate ø
s
equal


Similarly, by proving ΔAOD ≡ ΔCOB it can be shown that AD || BC
â ABCD is a parallelogram . . .
2 pairs of opp. sides are ||


In our sums, we may use ALL properties and theorem statements . . .
To prove that a quadrilateral is a ||
m
we may choose one of 5 ways:
1) Prove both pairs of opposite sides ||
(the definition)
.
2) Prove both pairs of opposite sides =
(a property)
.
3) Prove 1 pair of opposite sides = and ||
(a property)
.
4) Prove both pairs of opposite angles =
(a property)
. . . .
THE ANGLES

5) Prove that the diagonals bisect one another
(a property)
. . . .
THE DIAGONALS


Using diagonals . . .

To prove a parallelogram is a rectangle: prove that the diagonals are equal.
To prove a parallelogram is a rhombus: prove that the diagonals intersect
at right angles, or
prove that the diagonals bisect
the angles of the rhombus.
. . .
THE SIDES

A
B
C D
4
1
2
3
It doesn’t matter which
diagonal you draw

A
B
C D
3
1
2
4
>
>
A
B
C D
2
3
4
1
In Geometry, we never have
to repeat a 'logic sequence'
(as would’ve been required
here) –
we just say: Similarly, . . . ! TAS

15. 1 3

Copyright © The Answer Series


Area of Quadrilaterals and Triangles




A SUMMARY OF FORMULAE FOR
AREAS OF QUADRILATERALS

So far, we have established:
The
area of a trapezium =
1
2
(a + b).h

The
area of a kite =
1
2
the product of the diagonals




But also, remember:
The
area of a square = s
2
,

where s = the length of a side of a square

The
area of a rectangle =
ℓ %
b

where
ℓ
= the length & b = the breadth
The
area of a parallelogram = base % height






The
area of a rhombus
=
. . . . . . .
(1)
. . . . . . . . . . . . .


or =
. . . . . . .
(2)
. . . . . . . . . . . . .


Answers


(1) base % height (because a rhombus is a parallelogram), or
(2)
1
2
product of the diagonals (because a rhombus is a kite)
Why is the area of a parallelogram = base % height?




Compare ||
m
ABCD and rectangle EBCF


||
m
ABCD = ABCF + Δ2

& rectangle EBCF = ABCF + Δ1



But, we know that the opposite sides of rectangles and
parallelograms are equal

â Δ1  Δ 2 . . . SS90º or RHS
â Δ1 = Δ2 in area

â Area of ||
m
ABCD = Area of rectangle EBCF
= BC .FC . . .
= base % height (of the ||
m
)


ℓ
% b
b
ℓ
s
a
>
>
b
h
NB:
Study THE SUMMARY OF QUADRILATERALS for
'pathways of definitions, areas, properties', etc.
KNOW THIS WELL! (page 9)
See the explanation of
this formula below:

. . .

A

B C
D
E F
Δ1 Δ2 TAS

16. Copyright © The Answer Series
1 4


SOME IMPORTANT FURTHER FACTS ON AREAS
OF Δ
S
& QUADRILATERALS




1. Δ
s
on the same base & between the same || lines
are equal in area

Area of Δ =
1
2
base (BC) % height (h)

â ΔPBC = ΔABC = ΔQBC in area


2. Parallelograms on the same base and between
the same || lines are equal in area

Area of
||
m
= base (DC) % height (h)

â
||
m
ABCD =
||
m
PQCD in area


3. The median of a Δ bisects the area of the Δ

Area of ΔABD =
1
2
BD.h &

Area of ADC =
1
2
DC.h

But BD = DC . . .


â Area of ABD = Area of ADC [
1
2
Area of ABC!]

4. The diagonal of a parallelogram (or rhombus or
rectangle or square!) bisects the area

ΔABC ≡ ΔCDA . . . (SøS)

â ΔABC = ΔCDA [=
1
2

||
m
ABCD] in area


5. If two triangles lie between
the same parallel lines:


THE RATIO OF THEIR AREAS

=
THE RATIO OF THEIR BASES



area of ABD
area of DAC
Δ
Δ
=
1
2
1
2
..h
.4 .h
x
x
=
1
4
Also,
area of ABD
area of
Δ
ΔABC
=
1
2
1
2
..h
..h
x
x
5
=
1
5



base of ABD
= !
base of DAC
⎛⎞ Δ
⎜⎟
Δ
⎝⎠
â Area of ΔABD =
1
5
area of ΔABC

â Area of ΔABD =
1
4
Area of ΔDAC















The height of these Δ
s
& ||
ms
is the distance between the parallel lines . . .
A

B C D
4
x

x

h
x x
A

B C D
h
h
P
A

Q
B
C
h
P
A

Q
B
C
D
median AD bisects
the base

A
B
C DTAS

17. Copyright © The Answer Series

15

Gr 10: THE MIDPOINT THEOREM



FACT 1

The line segment through the midpoint of one side of a triangle, parallel to a second
side, bisects the third side.







FACT 2

The line segment joining the midpoints of two sides of a triangle is parallel to the
third side and equal to half of the third side.











PROOFS



1.

















2.










Given:
P & Q midpoints of AB & AC

Result:
PQ || BC & PQ =
1
2
BC
PQ
C B
A

Given:
P midpoint AB & PQ || BC

Result:
Q midpoint AC & PQ =
1
2
BC
P Q
C B
A
(See Exercise 4 Q3.2 for the proof) (See Exercise 4 Q3.1 for the proof)
Regard these Facts 1 & 2 as a special case
of the Proportion Theorem in Gr 12 Geometry.
Use the diagrams below to prove facts 1 and 2:
TAS

18. Grade 8 to 10

EUCLIDEAN GEOMETRY

EXERCISES & FULL SOLUTIONS


EXERCISE 1: Lines and Angles

EXERCISE 2: Lines, Angles and Triangles

EXERCISE 3: Quadrilaterals

EXERCISE 4: Midpoint Theorem
TAS

19. Copyright © The Answer Series
17
EXERCISE 1: Lines and Angles



(Answers on page 22)


1. 140º and 40º are adjacent supplementary angles:


The angle supplementary to
x
is:





2. In this figure : x
+ 90º + y = 180º . . .
ø
s
on a straight line




â
x
+ y = and so they are called angles.





3. Is ADB a straight line? Give a reason for your answer.











4.







5.
x
+ y = 180º . . .
ø
s
on a
straight line


and z + y = 180º . . .
ø
s
on a
straight line


â

6. Given: reflex
ˆ
AOD
= 200º
(see figure below)
.
Reasons:
Obtuse
ˆ
AOD
=


ˆ
BOD
=


ˆ
AOC
=



7.






ˆ
1
and
ˆ
2
are
ˆ
1
and
ˆ
2
are
ˆ
1
and
ˆ
2
are


ø
s

(NAME)
ø
s

(NAME)
ø
s

(NAME)




RELATIONSHIP : RELATIONSHIP : RELATIONSHIP :







8.
ˆ
1
and are corresponding ø
s
and they are



ˆ
1
and are alternate ø
s
and they are



ˆ
1
and are co-interior ø
s
and they are




9.
NB: It is ONLY BECAUSE THE LINES ARE , that
the corresponding and alternate ø
s
ARE EQUAL and the
co-interior ø
s
are SUPPLEMENTARY !!!


40º 140º
y
x

(2.1)
(2.2)
145º 45º
A D B
C
(3.1)
(3.2)
x
90º -
x
ADB
C
E
100º
40º
60º
?
(4.1)
x

?
(4.2)
The sum of the
adjacent angles
about a point
is 360º.
When two lines intersect,
the vertically opposite ø
s

are equal. Why?
(5)
O
A
C
B
D
200º
(6.1)
(6.3)
(6.4)
(6.5)
(6.6)
1
2
1
2
(7.1)
1
2
(7.2)
(7.3)
(7.4)
(7.5)
(7.6)
1
2
3
4
5
6
(8.1)
(8.2)
(8.3)
(8.4)
(8.5)
(8.6)
(9)
(1)
?
x
(6.2)
x

z
yTAS

20. Copyright © The Answer Series
18
EXERCISE 2:
Lines, Angles and Triangles



(Answers on page 22)

1. In the sketch,
AB is a straight line.
If
x
- y = 10º, find the
value of
x
and y.

2.1 Find the size of angles a to g (in that order),
giving reasons.








2.2 Calculate x
and give reasons.
2.2.1 2.2.2







2.3 If PQ = SR and
øPQR = øSRQ,
prove that
PR = SQ.





(Hint: first prove that ∆PQR ≡ ∆SRQ)

2.4 State whether PQ || RS,
giving reasons.






3. Determine the values of
x
and y in the following
diagrams. Give reasons for your answers.


3.1







3.2









4. Prove that
CP
=
ˆ
T

by first proving that
the 2 triangles
are congruent.





5. State whether the following pairs of triangles are
congruent or similar, giving reasons for your choice.

5.1







5.2









5.3






6.1 Find the value of
x
, by forming an equation first
and then solving for
x
.
Show all your working and give reasons.

6.1.1







6.1.2






6.2 In the diagram,
AB || ED and
BE = EC.
Also,
ˆ
ABE
= 27º
and
ˆ
BAC
= 53º.



6.2.1 Write down the sizes of
ˆ
BED
and
ˆ
CED
,
giving reasons.

6.2.2 Hence, or otherwise, calculate the
value of
x
, showing all working and
giving reasons.

7. Calculate the area
of the kite alongside.






x
+ y
y x

g
b
c d
a
e
f

35º 60º
3
x
4
x
5
x

PR
T W
Q S
76º
V
U 104º
B
AC
56º
37º
18
30
X
Z Y30
18
87º
N
69
L
M 75
51
H
2317
25
E
F
D
CB A
3
x

x
66º
A
DC
B
3
x
- 10º
x
+ 30º
F AC
E
G
BD
3
x
x

y
60º
2
x

120º
110º
x
PS
RQ
P
T
Q
R
S
x
x

A
B D C
53º
27º
x

E
4 cm
12 cm
13 cm
A
C
BD
E
K
T R
P
Q
A C
E
F
B
D
y
100º
60º
x
120º TAS

21. Copyright © The Answer Series
19
8. Calculate the value
of
x
giving
reasons.




9. If AC || DB, prove with
geometric reasons that
x
= y + z.




10. In ΔABC, BC is produced to D and CE || BA.
Prove that:

10.1
ˆˆˆ
1
A + B + C
= 180º
10.2
ˆ ˆˆ
ACD = A + B










11. Prove that
ˆ
DOC
= 45º.







12. If AB || CD,
ˆ
BOE
= 140º and
ˆ
AOC
= 35º.

Determine,
stating reasons,
the values of
x
and y.


13. In the accompanying
figure BG || CF.

Complete the
following statements
(giving reasons):



13.1
ˆˆˆ ˆ
1 + 2 + 3 + 4
= . . . . (degrees)
13.2
C
ˆ
1 + 11

= . . . . (degrees)
13.3
ˆˆˆ
6 + 7 + 2
= . . . . (degrees)
13.4
ˆˆ
9 + 6
= . . . (the no. of one angle)
13.5
ˆ
7
= . . . . (list only one angle)
13.6 If C
C
12 = 10
, then GD = . . . .
14.1 Express
ˆ
CEB
,
ˆ
ABE
and
ˆ
AEB
in terms of
x
.

14.2 What is the relationship
between CE and AD?
Explain.

14.3 If CE bisects
ˆ
ACD

what is the relationship
between BE and CD? Explain.

15.1 Make a neat copy of this sketch and fill in all the
other angles in terms of x. Reasons are
not required.








15.2 Complete the following statement:
∆ABE ||| ∆ . . . ||| ∆ . . .

15.3 If BC = 18 cm and BE = 12 cm,
calculate the length of

15.3.1 AE

15.3.2 AB correct to one decimal.

15.4 Hence calculate the area of rectangle ABCD.

16. In each of the following, state whether the given
triangles are congruent or not , and in each case
give a reason for your answer. Do not prove the
triangles congruent, but name each congruent
pair correctly.

16.1 16.2








16.3 16.4








16.5







17. Given ΔPQR and ΔABC.











17.1 Show that ΔPQR is NOT similar to ΔABC.
Clearly show all relevant calculations
and reasons.

17.2 Prove that ΔPQR is not right-angled.
A
2
x
+ 36º
BC D
3
x
- 50º
x

+ 10º A
C
y D
B
z
x
ABC D E
5
F
G
H
10
12
6789
2
31
4
11
A
E
DC
B
x

S
P
Q
R
A
B
C
D
E
B is the centre of the circle
B
AC D
B
AD
C
A

9
6
3
B
C
P
R
6
Q
8
12
E
A
B
C
D
y
y
x
x

A
B C D
E
2
31
BA
D
C
O
y
y
x
x

BE
D C
A
x
A B
CD
E
x


35º
2
1
3
140º
1
y
O TAS

22. Copyright © The Answer Series
20
EXERCISE 3: Quadrilaterals



(Answers on page 23)

1. Which of the following quadrilaterals is definitely a
parallelogram? (Not drawn to scale.)

A. B.



C. D.



2. If the area of
parallelogram PQRS
is equal to 90 cm
2

then PT is equal to
. . . cm.



3. In the given diagram,
PQRS is a parallelogram.
SP = PT and
UR = RQ.





Prove that TRUP is a parallelogram.

4. In the diagram,
ABCD is a
parallelogram.
AB is produced
to E and AD to F.

Prove that ECF
is a straight line.

5. ABCD is a rhombus.
By setting up an equation
and showing all steps and
reasons in the process,
find the value of y.
6. ABCD is a parallelogram.
E is a point on AD such
that AE = AB and
EC = CD.
ˆ
BEC
= 90º.

Calculate the size of
ˆ
EBC
.

7. In each of the following cases, calculate the value
of
x
giving reasons.

7.1 WXYZ is a
parallelogram.



7.2 ABCD is a square.

ˆ
BFD
= 125º.




8. Prove that ABCD is a parallelogram.







9. In the accompanying
figure ABCD is a
parallelogram.
The diagonal is
produced to E and
AB = BE and
AD = CE.

If
ˆ
CEB
=
x
, prove, giving reasons, that
ˆ
FDC
= 3
x
.

10. LM = 5 cm,
LP = 6 cm and
PS = 1 cm.



What type of quadrilateral is LMPN? Give reasons.
Find NM and MS.

11. ABCD is a parallelogram, and
AE and BF bisect
ˆ
A
and
ˆ
B

respectively.
P is the point of intersection
of AE and BF.

11.1 Find the magnitude
of
ˆ
APB
.

11.2 Show that BC = CF.

11.3 Prove that DF = EC.

12. ABCD is a parallelogram.
AP bisects ˆ
DAB
and
BP bisects
ˆ
ABC
.

Prove that AB = 2BC


13.1 Prove the theorem which says that if both pairs of
opposite angles of a quadrilateral are equal,
then it is a parallelogram.

13.2 PQRS is a quadrilateral
with PS = PR = QR and
PQ || SR.






Prove that it is a parallelogram.

13.3 Write down 2 facts about a rhombus which are
not generally true of any parallelogram.

14. PQRS is a parallelogram.







Prove that ABCD
is a rectangle.
P S
Q T R
15 cm
10 cm
P Q
S T R
U
AB
CD
4y - 18º
y
A
125º
x
B
CD
F
E
AB
C
D
E
x

F
A B
CD E F
x

y
x

P
y
BC
A DE
W
3
x
- 30º
x
+ 10º
X
Y Z
AB
CDP
1
2
2
1 3
1
2
P
S
C
Q
R
D
A
B
m
m
x

x

n
n
y
y
A
D F
C
E
B
P Q
S R
1
2
L
N
P
M
S
R
A
D
B F C
E
65º
65º
115º TAS

23. Copyright © The Answer Series
21
P
Q
R
S
W
T V
EXERCISE 4: Midpoint Theorem



(Answers on page 25)

1.1 Complete (giving missing words only):
The line segment joining the midpoints of two sides
of a triangle is . . . . . . to the third side and
equal to . . . . . . .

1.2 ΔABC has medians
BN and CM drawn,
cutting each other
at O.
P and Q are the
midpoints of BO
and CO
respectively.

Prove that MNQP is a parallelogram.

2. In the diagram below QS || TV; PQ || ST;
QT = TR = 9 cm and PS = 15 cm.








2.1 Prove VR = 7
1
2
cm.
2.2 Calculate PQ if PQ =
16
5
VR and hence
prove that
ˆ
PQR
= 90º.

2.3 Write down the length of ST.

3.1 With reference to the diagram,
prove the theorem which
states that PQ is parallel to DC
and half its length.



3.2 The diagram shows a
triangle PQR with M the
midpoint of PQ.
MNT is drawn parallel
to QR so that RT || QP.

Use this diagram to
prove the theorem
which states that
MN bisects PR.

4. Triangle DEF has G
the midpoint of DE,
H the midpoint of DF.
GH is joined.
HJ is parallel to DE.

Prove:
4.1 GHJE is a parallelogram.
4.2
ˆ
GDH
=
ˆ
JHF

4.3 JF = GH

5. M, N and T are the
midpoints of AB,
BC and AC in ∆ABC.
ˆ
A
= 60º and
ˆ
B
= 80º.

Calculate the angles
of ∆MNT.

6. ABCD is a square.
The diagonals AC and
BD intersect at O.
M is the midpoint of BO
and AM = ME.

6.1 Prove that MD || EC.

6.2 Prove that DOEC is
a parallelogram.

6.3 Determine, with reasons,
the size of:

6.3.1
ˆ
1
D

6.3.2
ˆ
ECD

7. With reference to the diagram,
prove that

7.1 FE = EC = CA
7.2 ΔBCA ≡ ΔAED









8. In ΔPQR, PT = TQ, while PV = VR = RS.












Show that WR =
1
4
QR.


9. ABCD is any quadrilateral.
E is a point BC. P, Q, R and S are the midpoints
of AB, AE, DE and DC.









Prove that:
9.1 PQ || RS
9.2 PQ + QR + RS =
1
2
(AD + BC)
P
S
Q T R
15
9
9 V
A
BC N
T
60º
M
80º
O
M
1
A
D
E
CB
A
D
P Q
C
D
G H
FJ E
A
B
P
Q
E
C
D
S
R
A
B C
M
O
N
P Q
P
M
T
R
N
Q
O
F
x
A
B
D
E
y
y
x
CTAS

24. Copyright © The Answer Series

22


ANSWERS TO EXERCISES

EXERCISE 1: Lines & Angles

1. 180º -
x


2.1 90º 2.2 complementary
3.1 No . . .
45º + 145º ≠ 180º

3.2 Yes . . .
90º – x + 90º + x = 180º

4.1 160º 4.2 360º -
x
5. â
x
= z

6.1 160º 6.2 revolution /
ø
s
about a point
6.3 20º 6.4
ø
s
on a straight line
6.5 20º
6.6
ø
s
on a straight line or vertically opposite to
ˆ
BOD


7.1 corresponding 7.2 co-interior 7.3 alternate
7.4 equal 7.5 supplementary 7.6 equal

8.1
ˆ
5
8.2 equal
8.3
ˆ
4
8.4 equal
8.5
ˆ
3
8.6 supplementary 9. parallel



EXERCISE 2:
Lines, Angles and Triangles


1. 2
x
+ 2y = 180º . . .
ø
s
on a straight line
â
x
+ y = 90

But
x
- y = 10º
â
x
= 50º and y = 40º . . .
by inspection



2.1 a = 60º . . .
vertically opposite angles

b = 35º . . .
alternate ø
s
; || lines

c = 35º . . .
ø
s
opp = sides

d = 85º . . .
sum of ø
s
in ∆

e = 60º - 35º = 25º . . .
ext. ø of ∆

f = 35º + 35º = 70º . . .
either ext. ø of ∆ or
corresp. ø
s
; || lines

g = 60º – 35º = 25º . . .
ext. ø of ∆


2.2.1 12
x
= 360º . . .
revolution

â
x
= 30º

2.2.2 120º + 110º +
x
= 2(180º)
â 230º +
x
= 360º
â
x
= 130º
2.2.3 In ∆
s
PQR & SRQ
[Note: order of letters!]

(1) PQ = SR . . .
(given)

(2)
ˆ
PQR
=
ˆ
SRQ
. . .
(given)

(3) QR is common

â ∆PQR ≡ ∆SRQ . . .
SøS

â PR = SQ

2.4
ˆ
PUV
= 180º - 76º = 104º . . .
ø
s
on str line

â
ˆ
PUV
=
ˆ
RVW

â Yes, PQ || RS . . .
because corresponding
ø
s
are equal


3.1
x
= 100º - 60º = 40º . . .
ext. ø of  EFC

y = 120º - 40º = 80º . . .
ext. ø of  ABC


3.2 4
x
= 2
x
+ 60º . . .
corr. ø
s
; BA || DC

â 2
x
= 60º
â
x
= 30º
y = 180º - 4
x
= 60º . . .
ø
s
on a straight line

4. In ∆
s
QRP and SRT
(1)
ˆ
Q
=
ˆ
S
(=
x
)
(2)
ˆ
QRP
=
ˆ
SRT
. . .
vertically opposite ø
s

(3) QR = SR . . .
given

â ∆QRP ≡ ∆SRT . . .
øøS

â
ˆ
P
=
ˆ
T


5.1 Congruent ; SøS . . .
[
ˆ
C
= 87º ø sum of ∆]


5.2 Similar ; sides in proportion
. . . [17:23:25 = 51:69:75]

5.3 Similar ; equiangular (alt.
ø
s

& vert. opp.
ø
s
)
. . . [no sides]

LOOKS MAY NOT DECEIVE !

6.1.1 3
x
= 66º +
x
. . .
ext ø of ∆

â 2
x
= 66º
â
x
= 33º

6.1.2 (3
x
- 10º) + (
x
+ 30º) = 180º . . .
co-int. ø
s
; AB || CD

â 4
x
+ 20º = 180º
â 4
x
= 160º
â
x
= 40º
6.2.1
ˆ
BED
= 27º . . .
alt. ø
s
; AB || ED


ˆ
CED
= 53º . . .
corresp. ø
s
; AB || ED


6.2.2
ˆ
ABD
=
x
. . .
corresp. ø
s
; AB || ED

â
ˆ
EBC
=
x
- 27º
â
ˆ
C
=
x
- 27º . . .
ø
s
opp = sides

â In ∆ABC : 53º +
x
+
x
- 27º = 180º . . .
sum of ø
s
in ∆

â 2
x
= 154º
â
x
= 77º

7. BE
⊥ AC
. . .
diagonals of a kite

BE = 5 cm . . .
5:12:13 ∆ ; Pythag.

Area of kite = 2 .
1
2
(12 + 4).5 . . .
2 % ∆

= 80 cm
2

[OR :
1
2
product of diagonals . . .
why?
]

8. 2
x
+ 36º = 3
x
- 50º +
x
+ 10º . . .
ext. ø of ∆

â - 2
x
= -76º
â
x
= 38º A VERY IMPORTANT THEOREM !

9.
ˆ
C
= z . . .
alt. ø
s
; AC || DB

â
x
= y + z . . .
ext. ø of ∆


10.1
ˆ
3
C
+
ˆ
2
C
+
ˆ
1
C
= 180º . . .
ø
s
on a straight line


But
ˆ
2
C
=
ˆ
A
. . .
alternate ø
s
; CE || BA

&
ˆ
3
C
=
ˆ
B
. . .
corresp. ø
s
; CE || BA


â
ˆ
B
+
ˆ
A
+
ˆ
1
C
= 180º

10.2
ˆ
ACD
= 180º -
ˆ
1
C
. . .
str. line

&
ˆ
A
+
ˆ
B
= 180º -
ˆ
1
C
. . .
ø sum of ∆

â
ˆ
ACD
=
ˆ
A
+
ˆ
B


11.
ˆ
DOC
=
x
+ y . . .
ext. ø of ∆


But 2
x
+ 2y = 90º . . .
ø sum of rt. ø
d
∆ABC

â
x
+ y = 45º

â
ˆ
DOC
= 45º

12.
ˆ
1
O
= 40º &
ˆ
3
O
= 40º . . .
ø
s
on a straight line

â
x
= 40º . . .
alt. or corresp. ø
s
; AB || CD


ˆ
1
C
= 35º . . .
alt. ø
s
; AB || CD

â y = 145º . . .
ø
s
on a straight line

. . .
2 prs. co-int. ø
s
;

||
lines TAS

25. Copyright © The Answer
23

13.1 360º . . .
revolution

13.2 180º . . .
co-interior ø
s
; BG || CF

13.3 180º . . .
vert. opp. ø
s
; ø sum of ∆

13.4
C12
. . .
ext. ø of ∆

13.5
ˆ
9
. . .
corresp. ø
s
; BG || CF

13.6 BD . . .
C12

=
C10
C
C11

=
C9
; sides opp = ø
s



14.1
ˆ
CEB
=
x
. . .
ø
s
opp = sides


ˆ
ABE
= 2
x
. . .
ext. ø of ∆


ˆˆ
A + AEB
= (180º - 2
x
) . . .
sum of ø
s
in ∆

â
ˆ
AEB
= 90º -
x
. . .
ø
s
opp = sides







14.2 They are perpendicular to each other; i.e. CE ⊥ AD;
Explanation :
ˆ
CEB
and
ˆ
AEB
are complementary
. . .
x + (90º - x) = ?

14.3 They are parallel; i.e. BE || CD ;
Explanation :
ˆ
ECD
=
x
. . .
CE bisects
ˆ
A
CD

â
ˆ
ECD
= alternate angle,
ˆ
CEB


15.1







15.2 ∆ABE ||| ∆ECB ||| ∆DEC

15.3.1 ∆ABE ||| ∆ECB
C AE
BE
=
BE
BC
. . .
sides in proportion

%BE) â AE =
22
BE 12
= = 8 cm
BC 18


15.3.2 AB
2
= 12
2
- 8
2
= 80 . . .
Theorem of Pythagoras

â AB =
80

j 8,9 cm

15.4 Area of rect. ABCD = 8,9 % 18 j 161 cm
2

16.1 Yes, ∆ABC ≡ ∆EDC . . .
øøS


16.2 No . . .
2 sides and an angle, but the angle isn’t included


16.3 Yes, ∆ABD ≡ ∆ACD . . .
RHS



16.4 Yes, ∆ADB ≡ ∆CDB . . .
SSS [Note: equal radii]


16.5 No . . .
2 angles and a side, but they don’t correspond


17.1
12 4 8 4 6
=; = but =2
93 63 3

â The sides are not in proportion
â ∆PQR ||| ∆ABC

17.2 12
2
= 144 and 6
2
+ 8
2
= 36 + 64 = 100
â 12
2
g 6
2
+ 8
2

â
ˆ
Q
g 90º, i.e. ∆PQR is not rt. ø
d
. . .


EXERCISE 3: Quadrilaterals


1. D : diagonals bisect one another

2. Area of ||
m
PQRS = 15 % PT = 90 . . .
base % height

â PT = 6 cm

3.





Let
ˆ
S
=
x

Then
ˆ
PTS
=
x
. . .
ø
s
opp = sides

â
ˆ
TPQ
=
x
. . .
alt. ø
s
; PQ || SR in ||
m
PQRS


But
ˆ
Q
=
x
. . .
opp. ø
s
of ||
m
PQRS

â
ˆ
RUQ
=
x
. . .
ø
s
opp = sides

â
ˆ
TPQ
=
ˆ
RUQ

â PT || UR . . .
corresponding ø
s
equal


But PU || TR . . .
opposite sides of ||
m
PQRS

â TRUP is a parallelogram . . .
both pairs opp. sides ||






4.





Let
ˆ
F
=
x

then
ˆ
BCE
=
x
. . .
corr. ø
s
; BC || AD(F) in ||
m


Similarly, let
ˆ
E
= y
then
ˆ
DCF
= y

In ΔAEF :
ˆ
A
+
x
+ y = 180º . . .
ø sum of 

But
ˆ
A
=
ˆ
BCD
. . .
opposite ø
s
of ||
m

â
ˆ
BCD
+
x
+ y = 180º
â ECF is a straight line . . .
conv. of 'ø
s
on a str. line'


5.
ˆ
DAC
= y . . .
diagonals of a
rhombus bisect the
ø
s
of the rhombus

â
ˆ
DCA
= y . . .
ø
s
opp = sides,
sides of a
rhombus (or alternate angles)


â 4y - 18º + 2y = 180º . . .
ø sum of 

â 6y = 198º
â y = 33º

6.
ˆx
1
=
ˆx
2
. . .
ø
s
opp = sides

=
ˆx
3
. . .
alt. ø
s
; AD || BC in ||
m

(=
x
, say)



Similarly, y
1
= y
2
= y
3
(= y, say)


x
+ y = 90º . . .
str.

ˆ
A
ED

or ø
s
of BEC

and y = 2
x
. . .
opp. ø
s
of ||
m

â 3
x
= 90º
â
x
= 30º, i.e.
ˆ
EBC
= 30º

7.1 3
x
- 30º =
x
+ 10º . . .
opp. ø
s
of a ||
m

â 2
x
= 40º
â
x
= 20º

BE
D C
A
x
2
x

x
P Q
S
T
R U
x
x

x
x

x
A B
CD
4y - 18º
y
y
y
W
3
x
- 30º
x
+ 10º
X
Y Z
A D
x
1x
3
x
2
y
2
y
1
y
3
B C
E
The angle is not
included, but
it is a right angle
CONVERSE
of Theorem
of Pythag.
A
E
D C
B
x

90º -
x

12
x

90 º-
x

90º -
x

18
x

N
B: The order of
the letters!
A D F
C
E
B
y
y
x

x
TAS

26. Copyright © The Answer Series

24

7.2
ˆ
BAE
= 45º . . .
diagonals of a square bisect
(right) angles of a square

â
ˆ
ABE
=
x
- 45º . . .
ext. ø of
Δ


ˆ
FAB
= 90º . . .
ø of square



ˆ
BFD
=
ˆ
ABE
+
ˆ
FAB
. . .
ext. ø of Δ

â 125º =
x
- 45º + 90º
â
x
= 80º


8.







ˆ
AEC
+
ˆ
FCE
= 115º + 65º = 180º
â AE(D) || (B)FC . . .
co-interior

ø
s
supplementary



ˆ
B
= 65º . . .
ø
s
opp = sides


&
ˆ
CED
= 65º . . .
str. line AED

â
ˆ
D
= 65º . . .
ø
s
opp = sides

â
ˆ
ECD
= 50º . . .
ø sum of Δ


â
ˆ
B
+
ˆ
DCB
= 65º + 50º + 65º = 180º
â AB || DC . . .
co-interior

ø
s
supplementary

â ABCD is a ||
m
. . .
2 pairs opp. sides ||




9.







ˆ
EAB
=
x
. . .
ø
s
opp = sides

â
ˆ
DCA
=
x
. . .
alt. ø
s
; DC || AB in ||
m



ˆ
CBE
=
x
. . .
ø
s
opp = sides

â
ˆ
ACB
= 2
x
. . .
ext. ø of Δ
â
ˆ
DAC
= 2
x
. . .
alt. ø
s
; AD || BC in ||
m



ˆ
FDC
=
ˆ
DAC
+
ˆ
DCA
. . .
ext. ø of Δ
= 2
x
+
x
= 3
x


10. A rhombus ; 2 prs. opp. sides ||
and diagonals intersect
at right
ø
s





LR = RP = 3 cm . . .
diagonals bisect

â RM = 4 cm . . .

ˆ
L
RM
= 90º ; 3:4:5∆ ; Pyth.
â NM = 8 cm . . .
diagonals bisect

MP = 5 cm . . .
sides of rhombus
â MS
2
= 5
2
- 1
2
= 24 . . .
Pythagoras

â MS =
24
j 4,9 cm



11.1







2
x
+ 2y = 180º . . .
co-int. ø
s
; AD || BC in ||
m

â
x
+ y = 90º
â
ˆ
APB
= 90º . . .
ø sum of 


11.2
ˆ
BFC
=
ˆ
ABF
= y . . .
alt. ø
s
; AB || DC

â
ˆ
CBF
=
ˆ
CFB

â BC = CF . . .
sides

opp = ø
s



11.3 DF = DE - FE
& EC = CF - FE


ˆ
AED
=
ˆ
BAE
(=
x
) . . .
alt. ø
s
; AB || DC in ||
m

â DE = AD . . .
ø
s
opp = sides

= BC . . .
opp. sides of ||
m

= CF . . .
proved in 11.2

â DF = EC



12.







ˆ
1
P
=
ˆ
2
A
. . .
alt. ø
s
; AB || DC in ||
m

=
ˆ
1
A
(=
x
) . . .
given

â DP = DA . . .
ø
s
opp = sides


Similarly,
ˆ
3
P
=
ˆ
1
B
=
ˆ
2
B
= y
â CP = BC

But BC = DA . . .
opp. sides of ||
m

& AB = DC . . .
opp. sides of ||
m

= DP + CP
= DA + BC
= 2BC


13.1 In the notes (Theorem 4)

13.2 PQ || SR

Let
ˆ
S
=
x
; then

ˆ
PRS
=
x

. . .
ø
s
opp = sides

â
ˆ
2
P
=
x

. . .
alt. ø
s
;
PQ || SR

â
ˆ
Q
=
x

. . .
ø
s
opp = sides


â In ∆
s
PSR and PRQ

ˆ
1
P
and
ˆ
PRQ
= 180º - 2
x
. . .
ø sum of 
s

â PS || QR . . .
alt. ø
s
equal

â PQRS is a ||
m
. . .
2 prs. opp. sides ||


OR: Could've proved 2 prs. opp.
ø
s

equal
OR : Could've proved 2 prs. opp. sides equal

13.3 The diagonals intersect at right angles.
The diagonals bisect the angles of the rhombus.
(& 2 adjacent sides are equal)



L
N P
M
S
R
5
3
3
1
A B
C D EF
x

y
x

P
y
y
x
A B
C
D
E
F
2
x
2
x

x

x

x
x

x

x
y
y
y
A B
C D

P

1
2
2
1 3
1
2
P Q
SR
1
2
x

x

x

x
A
125º
x

B
CD
F
E
65º
A D
B F C
E
65º
65º
115º
65º
65º
50ºTAS

27. Copyright © The Answer
25

14.








2m + 2n = 180º . . .
co-int. ø
s
; PQ || SR in ||
m

â m + n = 90º
â
ˆ
PAS
= 90º . . .
ø sum of 

â
ˆ
BAD
= 90º . . .
vertically opp ø
s



Similarly,
x
+ y = 90º
â
ˆ
QCR
= 90º
â
ˆ
BCD
= 90º

So, too, n + y = 90º
â
ˆ
B
= 90º . . .
ø sum of 


â The 4
th
angle,
ˆ
D
= 90º . . .
ø sum of quad.


â ABCD is a rectangle . .
. all angles = 90º




EXERCISE 4: Midpoint Theorem


1.1 . . . parallel . . . half of the third side.

1.2





Join MN and PQ
In ∆ABC : M & N are midpoints of AB & AC
â MN || BC and MN =
1
2
BC . . .
midpoint thm

& In ∆OBC : P & Q are midpoints of OB & OC
â PQ || BC and PQ =
1
2
BC . . .
midpoint thm

â MN || PQ and MN = PQ . . .
both || and =
1
2
BC

â MNQP is a ||
m
. . .
1 pair of opp. sides = and ||

2.1 In ∆RQP : T midpoint QR & TS || QP

â S midpoint PR . . .

â SR = 15 cm

Similarly : In ∆RQS : V midpoint SR
â VR =
1
2
(15) =
1
7 cm
2


2.2 PQ =
16 15
=24 cm
52
%

â sides of ∆PQR :
18:24:30 = 3:4:5
â
ˆ
PQR
= 90º . . .
ratio of sides = Pythag. 'triple'
i.e. converse Pythag.


2.3 ST =
1
2
(24 cm) = 12 cm . . .


3.1 Construction :
Extend PQ to R
such that PQ = QR
Join AR, PC and CR

Proof : APCR is a ||
m
. . .
diagonals bisect one another

â CR = and || AP
â CR = and || PD
â PDCR is a ||
m
. . . 1 pr. opp. sides = and ||
â PR || and = DC
â PQ || and =
1
2
DC . . .
PQ =
1
2
PR


3.2. MQRT is a ||
m
. . .
â TR = and || MQ
â TR = and || PM
â In ∆
s
PNM & RNT

(1) PM = RT . . .
proved

(2)
ˆ
PNM
=
ˆ
RNT
. . .
vertically opp

ø
s


(3)
ˆ
PMN
=
ˆ
NTR
. . .
alternate

ø
s
; PM || TR

â ∆PNM ≡ ∆RNT . . .
øøS

â PN = NR
4.1




HJ || GE . . .
given

& In ∆DEF : G & H are midpoints of DE & DF
â GH || EJ(F) [and GH =
1
2
EF]
â GHJE is a ||
m
. . .
2 pairs opp. sides ||


4.2
ˆ
GDH
=
ˆ
JHF
. . .
corresp. ø
s
; HJ || DE

4.3 EJ = GH . . .
opp. sides of ||
m

=
1
2
EF . . .
see 4.1

( â JF =
1
2
EF too)
â JF = GH


5.







ˆ
C
= 180º - (80º + 60º) . . .
sum of ø
s
in 

= 40º


In ΔABC : M & T are midpoints of AB and AC
â MT || BC . . .
midpoint thm
.
â
ˆ
AMT
= 80º . . .
corresp. ø
s
; || lines



Similarly, M & N are midpoints of AB & BC
â
ˆ
BMN
= 60º . . .
corresp. ø
s
; MN || AC

â
ˆ
TMN
= 180º - (80º + 60º) . . .
ø
s
on a str. line

= 40º

The same method can be followed to determine the
other two angles of ΔMNT.

Answer : 40º ; 80º ; 60º
OR : Pythag
9:12:15
= 3:4:5 !
midpoin
t

theorem

converse of
mid
p
oin
t
theorem
A
BC
M
O
N
P
Q
A
D
P
Q
C
R A
BC N
T

60º
M
80º
P
S
C
Q
R
D
A
B
m
m
x

x
n
n
y
y
P
S
Q
T
R
15
9
9 V
2 pairs
o
pp
. sides
||
D
GH
F J E
P
M
T

R
N
Q TAS

28. Copyright © The Answer Series

26

P
Q
R
S
W
T

V
6.1 In ∆AEC : O midpoint AC . . .
diagonals bisect

& M midpt. AE . . .
given


â MO(D) || EC . . .
midpoint theorem



6.2 We have DO || CE . . .
in 6.1

& DO = OB . . .

= 2OM . . .
M midpt. BO

= EC . . .
midpoint. theorem


â DOEC is a ||
m
. . .
1 pair of opp. sides = and ||


6.3.1
ˆ
1
D
= 45º . . .
right ø
s
of sq. bisected by diagonal

6.3.2
ˆ
ECA
=
ˆ
BOA
. . .
corresp.

ø
s
; MO || EC in 6.1

= 90º . . .
diagonals of sq. int. at rt.

ø
s

&
ˆ
OCD
= 45º . . .
just as
ˆ
1
D
above

â
ˆ
ECD
= 90º + 45º = 135º


7.1 In ∆FBC : D midpoint FB & DE || BC
â FE = EC . . .
converse of midpoint thm



& In ∆EDA : O midpoint AD and (B)OC || DE
â EC = CA . . .
converse of midpoint thm



â FE = EC = CA




7.2 In ∆
s
BCA & AED

(1)
ˆ
CBA
=
ˆ
EAD
. . .
both = x

(2)
ˆ
BCA
=
ˆ
AED
. . .
corresp. ø
s
; DE || BC

(3) CA = FE . . .
in 7.1

= ED . . .
ø
s
opp = sides


â ∆BCA ≡ AED . . .
øøS



8. In ∆PQR : T & V are midpoints of PQ & PR
â TV || QR and =
1
2
QR . . .

â In ∆STV : R midpoint VS and WR || TV
â WR =
1
2
TV
=
1
2
(
1
2
QR)
=
1
4
QR









9.



9.1 In ΔABE :
P & Q are midpoints of AB & AE
â PQ || BE(C)

Similarly, in ΔDEC :
RS || (B)EC
â PQ || RS
. . .
both are
parallel
to BEC


9.2 In Δ
s
ABE, AED and DEC :
PQ + QR + RS =
1
2
BE +
1
2
AD +
1
2
EC
=
1
2
(AD + BE + EC)
=
1
2
(AD + BC)















































midpoin
t

theorem

These Geometry materials (Booklets 1 to 4)
were created and produced by
The Answer Series Educational Publishers (Pty) (Ltd)
to support the teaching and learning of
Geometry in high schools in South Africa.

They are freely available to anyone
who wishes to use them.

This material may not be sold (via any channel)
or used for profit-making of any kind.
This drawing looks confusing at first. But, look at each
triangle separately – the 'middle' one is just upside down!
– and apply the facts to each, one at a time.
A
B
P
Q
E
C
D
S
R
diagonals of

s
q
. bisec
t

O
M
1
A D
E
CB
O
F
x

A
B
D
E
y

y

x
C TAS