6.2 Axioms, Postulates
and Enunciations on lines:
6.2.1 Axioms
In geometry we have to accept some facts, without
discussion and proof. They are called axioms and postulates. Axiom means
‘Saying’
as per English dictionary. Axioms are
required more in Geometry than in other branches of mathematics. They are self evident truths
formed as a result of
observations and intuition.
Definition:
‘Axiom’ is a statement, the
truth of which is accepted without any proof.
Notes :
1.
Axioms should not contradict each other. They must be
consistent
2.
Axioms should be independent(No axiom should be created
based on another axiom)
3.
Axioms should be minimum in
numbers.
1. In Algebra, we know that if
a=b and b=c then a=c


In
Figure 1, length of AB=3cm and in Figure 2, length of CD=3cm, then we say
that AB=CD.. 
In
Figure 3, _{}ABC = 50^{0} and in Figure 4, _{}PQR = 50^{0}, then we say
that _{}ABC = _{}PQR. 
In
Geometry this property is stated as an axiom as follows: 6.2.1 Axiom 1: Things which are equal to the same thing are equal to
one another. 
2. In Algebra, we know
that if a=b then a+c=b+c for any c


In
Figure 5, AB=3cm and BE=2cm. In Figure 6, CD=3cm and DF =2cm. If
BE and DF are added to AB and CD respectively then we have AE=AB+BE=5cms and
CF=CD+DF=5cm and we say AE=CF 
In Figure 7, _{}ABC = 20^{0 }and _{}CBD = 40^{0}. In Figure 8, _{}PQR = 20^{0 }and _{}RQS = 40^{0}.^{} If _{}CBD and _{}RQS are added to _{}ABC and _{}PQR respectively then we have_{}ABD= 60^{0} and _{}PQS = 60^{0} and we say _{}ABD =_{}PQS.^{} 
In
Geometry this property is stated as an axiom as follows:
6.2.1 Axiom 2: If equals are added
to equals then the result is also equal 
3. In Algebra, we know that if a=b then ac=bc


In
Figure 9, AE=5cm and BE=2cm. In Figure 10, CF=5cm and DF =2cm. If
BE and DF are subtracted from AE and CF respectively then we have
AB=AEBE=3cms and CD=CFDF=3cm and we say AB=CD. 
In Figure 11, _{}ABD = 60^{0 }and _{}CBD = 40^{0}. In Figure 12, _{}PQS = 60^{0 }and _{}RQS = 40^{0}.^{} If _{}CBD and _{}RQS are subtracted from _{}ABD and _{}PQS respectively then we have_{}ABC = 20^{0} and_{}PQR = 20^{0} and we say _{}ABC =_{}PQR. 
In
Geometry this property is stated as an axiom as follows:
6.2.1 Axiom 3: If equals are
subtracted from equals then the result is also equal 
4. In
Algebra, we know that if n >1 then a
> a/n


In Figure 13, compare AB with AE and EB, We say that AB>AE and AB>BE. 
In Figure 14, compare _{}ABC with _{}ABD and _{}DBC. We say that _{}ABC >_{}ABD and _{}ABC >_{}DBC. 
In
Geometry this property is stated as an axiom as follows: 6.2.1 Axiom 4: The whole is
greater than its parts 
Note: These axioms are used in any branch of Mathematics and
not necessarily only in Geometry.
6.2.2 Postulates:
English dictionary gives the meaning for postulate
as ‘guess’ or ‘propose’ or ‘assume’.
Definition: ‘Postulates’ are mathematical
statements in geometry, which are
assumed to be true without proof.
They are like axioms but they can be cross verified
by actual construction and measurements.
Have
you ever thought about how people mark the lanes for 100 meters running race? They
identify two points which are 100 meters apart and draw the line joining
these two points with chalk powder. Are they using any rule of geometry? 


In
the adjacent figure, A and B are two points and AB is the line passing
through them. 


This
property is stated as a postulate as follows: 6.2.2 Postulate 1: One and only
one line can be drawn passing through two points. Is it not interesting to note that without knowing this postulate, lanes/tracks are marked by workmen? 

Have
you thought of any assumption that is made when spokes of the bicycle are
joined at the center of the wheel? In
the adjacent figure O is the center of the wheel and several lines (spokes)
pass through it. 


This
property is stated as a postulate as follows: 6.2.2 Postulate 2: Any number of
lines can be drawn passing through a point. Spokes
in the wheel of a bullock cart/bicycle is a best example of use of this
postulate. 

In
the adjacent figure PQ is a line which is extended on both sides. 


This
property is stated as a postulate as follows: 6.2.2 Postulate 3: A straight line can be extended to any length on both sides 

In
the adjacent figure, two rays are drawn in opposite directions from O (OA and



This
property is stated as a postulate as follows: 6.2.2 Postulate 4: Angle formed at the common point of two opposite rays
is 180^{0}. 

In
how many ways can you sit on the chair? Either you sit with legs crossed or with
legs parallel to each other…
In
the adjacent figure the lines AB and CD cross at O or they are parallel to
each other. 


This
property is stated as a postulate as follows: 6.2.2 Postulate 5: Two
straight lines either meet at only one point or they do not have a common
point. 

Have
you ever thought of what happens if railway lines meet? Journey by Railways
will not be possible at all… In
the adjacent figure, lines AB and CD which are parallel, will never meet even
if they are extended on both sides (like railway lines). 


This
property is stated as a postulate as follows: 6.2.2 Postulate 6: Two
parallel lines in a plane never meet even if produced infinitely on either
side. 

Conclusion: From postulates 5 and
6, we conclude that any two straight lines which do not have a common point are
parallel to each other.
6.2.3 Enunciations/Rules
Let us learn what Enunciations are. The English
dictionary gives the meaning of Enunciations as ‘guess’ or ‘propose’.
Enunciations are true facts which can only be
verified by construction or measurements.
In the adjacent figure _{}AOC + _{}COB = 180^{0} 

This
property is stated as an Enunciation as follows: 6.2.3 Enunciation 1: If a ray stands on a straight line, the sum of angles
formed at the common point is 180^{0}. This
is some time refereed as ‘Linear pair axiom’ 

In
the adjacent figure, the lines AB and CD intersect at O. we notice _{}AOC = _{}DOB and _{}AOD = _{}COB 


This
property is stated as an Enunciation as follows: 6.2.3 Enunciation 2: If two straight lines intersect, the vertically
opposite angles are equal. The
good example is angles formed in a scissor 


The above enunciation can also be
proved as follows:
No 
Statement 
Reason 

1 
_{}AOC + _{}COB = 180^{0} 
Enunciation 1: AB is a straight
line and OC is standing on AB. 

2 
_{}DOA + _{}AOC = 180^{0} 
Enunciation 1: DC is a straight
line and OA is standing on
DC. 

3 
_{}AOC + _{}COB =_{}DOA + _{}AOC 
Axiom 1 

4 
_{}COB = _{}DOA 
Axiom 3(equal thing subtracted
is _{}AOC) 
Similarly we can prove _{}AOC = _{}DOB.
Definitions:
1.
Two angles are said to be ‘adjacent’ if
they have a common side and a common end point. (In Figure 1, _{}ABC and _{}CBD are adjacent angles because
they have C as common point and BC as common side) 

2.
Two angles are said to be ‘complimentary’,
if the sum of their measures is 90^{0}(In Figure 2, _{}PQS and _{}SQR are complimentary angles
because PQR is a right angle and _{}PQR=90^{0}) 

3.
Two angles are said to be ‘supplementary’,
if the sum of their measures is 180^{0}(In Figure 3, _{}XOZ and _{}ZOY are supplementary angles
because OZ is a ray standing on the straight line XOY and _{}XOY = 180^{0}) 

4.
A line which intersects two or more coplanar (on the same plane) lines at different
points is called ‘transversal’ (In Figure 4, AB and CD are coplanar
lines. EF is the line which cuts AB and CD at G and H respectively. EF is
called transversal.) The
length of the transversal between parallel lines is called ‘intercept’ In Figure 4, GH is an intercept. 

Examples of different types of angles are:
Adjacent Angles 
Vertically Opposite Angles(4
pairs) 
Alternate Angles (2 pairs) 
Corresponding Angles(4 pairs) 
Interior Angles (2 pairs) 

_{}AGE and _{}EGB 
_{}AGE and _{}HGB 
_{}AGH and _{}GHD 
_{}EGB and _{}GHD 
_{}AGH and _{}GHC 

_{}CHF and _{}FHD 
_{}CHF and _{}GHD 
_{}BGH and _{}CHG 
_{}AGE and _{}CHG 
_{}BGH and _{}GHD 

……. 
_{}AGH and _{}EGB 

_{}AGH and _{}CHF 



_{}CHG and _{}FHD 

_{}BGH and _{}DHF 

In
the adjacent figure, AB and CD are parallel lines and EF is a transversal.
Then _{}EGB = _{}GHD, _{}AGH = _{}CHF, _{}AGE =_{}GHC and _{}BGH = _{}DHF. 

This
property is stated as an Enunciation as follows: 6.2.3 Enunciation 3: If a transversal cuts two parallel lines, then the
corresponding angles are equal. 
6.2.3 Enunciation 4: If a transversal cuts
two lines in such a way that the corresponding angles are equal then the two
lines are parallel.
(This is converse of Enunciation 3.)
6.2 Problem
1 : In the figure, O
is a point on the straight line AB, Line OP stands on AB at O. OQ bisects _{}POB and OR bisects _{}AOP
Prove that _{}ROQ = 90^{0}.
Soluition :
No 
Statement 
Reason 

1 
_{}POQ = _{}QOB 
OQ bisects _{}POB. 

2 
_{}POB = 2* _{}POQ 
Step 1 

3 
_{}AOP =2*_{}ROP 
OR bisects _{}AOP 

4 
_{}AOP + _{}POB = 180^{0} 
Enunciation 1: AB is a straight
line and OP is standing on AB 

5 
2* _{}ROP +2*_{}POQ =180^{0} 
Step 4,2,3 

6 
2(_{}ROP +_{}POQ) =180^{0} 
Simplification 

7 
_{}ROP +_{}POQ =90^{0} 


8 
_{}ROQ=90^{0} 
_{}ROP +_{}POQ=_{}ROQ 
6.2 Problem 2: In the figure, O
is a point on the straight line AB. Lines OP and OQ stand on AB at O. Find all
the angles and show
that _{}QOP is a right angle.
Soluition :
No 
Statement 
Reason 

1 
AOP+POB=180^{0} 
Enunciation 1: OP is standing on
AB 

2 
x+2x = 180^{0} i.e.3x =180^{0 } i.e. x =60^{0} 
Substitution, Simplification 

3 
AOQ+QOB=180^{0} 
Enunciation 1: OQ is standing on
AB 

4 
y+5y = 180^{0
} i.e. 6y = 180^{0 } i.e.
y =30^{0} 
Substitution, Simplification 

5 
_{}AOP = x = 60^{0 }_{}POB = 2x = 120^{0 } ^{} 
Substitution 

6 
_{}AOQ = y = 30^{0 },_{}QOB = 5y =150^{0} 
Substitution 

7 
_{}QOP = _{}QOA +_{}AOP= y+x
=30^{0 }+ 60^{0 }= 90^{0 } 
Substitution 
6.2 Problem 3: In the figure, O
is a point on the straight line AB. If ab =80^{0}, find all angles.
Solution:
No 
Statement 
Reason 

1 
_{}AOP+_{}POB=180^{0} 
Enunciation 1: OP is standing on
AB 

2 
a+b= 180^{0} 
Substitution 

3 
b = 180^{0}a 
By transposition 

4 
ab = 80^{0} 
Given 

5 
ab= a – (180^{0} a) =
2a 180^{0} 
Substitute value of b in step 4 

6 
2a 180^{0}=80^{0} 
Equating 4 and 5 

7 
2a =80^{0}+180^{0}=
260^{0}: 2a =260^{0} 
By transposition, Simplification 

8 
a= 130^{0}:b =50^{0} 
Simplification, Substitution 
6.2 Problem 4: In the figure,
Solution:
No 
Statement 
Reason 

1 
_{}AOP+_{}POB=180^{0} 
Enunciation 1: OP is standing on
AB 

2 
_{}AOP = 180^{0}_{}POB 
By transposition 

3 
_{}POB = _{}BOQ 
Given that 

4 
_{}AOP = 180^{0}_{}BOQ 
Substitute 3 in 2. 

5 
_{}AOQ+_{}QOB=180^{0} 
Enunciation 1: OQ is standing on
AB 

6 
_{}AOQ = 180^{0}_{}BOQ 
By transposition 

7 
_{}AOP = 180^{0}_{}BOQ= _{}AOQ 
Equating Step 4 and 6 
6.2 Problem 5: In the figure, PQ
and RS are straight lines. OA bisects _{}POR and
Solution:
No 
Statement 
Reason 

1 
_{}POR = 2_{}AOP 
Given that OA bisects _{}POR 

2 
_{}SOQ = 2_{}BOQ 
Given that 

3 
_{}POR = _{}SOQ 
Enunciation 2, vertically
opposite angles are equal: PQ and RS are straight lines 

4 
2_{}AOP = 2_{}BOQ: _{}AOP =_{}BOQ 
Substituting 1 and 2 in 3 

5 
_{}AOB = _{}AOP+_{}POS+_{}SOB 


6 
= _{}BOQ+_{}POS+_{}SOB 
Substituting _{}BOQ for _{}AOP from of step 4 

7 
=_{}POS+_{}SOB+_{}BOQ 
Re arranging angles 

8 
= 180^{0} 
PQ is a straight line and OS is
ray on that line and _{}SOQ =_{}SOB+_{}BOQ. AB is a straight line 
6.2
Problem 6: In the figure , _{}ABC = _{}ACB Prove that _{}ACQ =_{}ABP and _{}CBR =_{}BCS
Solution:
No 
Statement 
Reason 

1 
_{}ACB+_{}ACQ = 180^{0} 
BC is a straight line. 

2 
_{}ACB = 180^{0} – _{}ACQ 
By transposition 

3 
_{}PBA+_{}ABC = 180^{0} 
BC is a straight line. 

4 
_{}PBA = 180^{0} – _{}ABC 
By transposition 

5 
= 180^{0} – _{}ACB^{} 
it is given that _{}ABC=_{}ACB 

6 
= 180^{0} – (180^{0}
–_{}ACQ) 
Substitute for _{}ACB from step 2 

7 
= _{}ACQ 
Simplification ( this proves
first part) 

8 
_{}PBR=_{}ABC 
They are opposite angles 

9 
_{}QCS=_{}ACB 
They are opposite angles 

10 
_{}PBR = _{}QCS 
From step 8 and 9 and it is
given that _{}ABC=_{}ACB 

11 
_{}CBR = 180^{0} – _{}PBR 
_{}CBR +_{}PBR = 180^{0} 

12 
= 180^{0} – _{}QCS 
_{}PBR =_{}QCS from step 10 

13 
=180^{0} – (180^{0}
– _{}BCS) 
_{}QCS+_{}BCS = 180^{0} 

14 
= _{}BCS 
Simplification (this proves
second part) 
6.2
Problem 7: In the figure, _{}AGE=120^{0} and _{}CHF = 60^{0}. Find out
whether ABCD or not.
_{}EGB and _{}GHD
Solution:
No 
Statement 
Reason 

1 
_{}CHF = _{}GHD = 60^{0} 
_{}CHF and _{}GHD are vertically opposite
angles and it is given that _{}CHF = 60^{0}. 

2 
_{}EGB = 60^{0}. 
Angles on the straight line _{}AGE + _{}EGB =180^{0} and it is given that _{}AGE = 120^{0} 

3 
_{}EGB = _{}GHD 
Step
1 and 2 


Since
_{}EGB and _{}GHD are corresponding angles and
are equal, by Enunciation 4, AB and CD are parallel. 
6.2 Problem 8: In the figure,
ABCD, EF cuts them at G and H respectively. If _{}AGE and _{}EGB are in the ratio of 3:2,
find all the angles in the
figure.
Solution:
Since AB is a straight line, _{}AGE + _{}EGB =180^{0}.
Since the ratio of angles is 3:2, 180^{0}
needs to be split in to two angles as per this ratio. Total parts = 3+2 =5.
5 parts = 180^{0}
_{}1 part = 180^{0}/5 = 36^{0 }
_{}_{}AGE = 3parts = 3*36^{0} =
108^{0}
_{}_{}EGB = 2parts = 2*36^{0} = 72^{0}
No 
Statement 
Reason 

1 
_{}AGE =_{} HGB =108^{0} 
Vertically opposite angles 

2 
_{}EGB = _{}AGH = 72^{0} 
Vertically opposite angles 

3 
_{}EGB = _{}GHD = 72^{0} 
Corresponding angles 

4 
_{}AGE = _{}CHG =108^{0} 
Corresponding angles 

5 
_{}DHF = _{}CHG =108^{0} 
Vertically opposite angles 

6 
_{}CHF = _{}GHD = 72^{0} 
Vertically opposite angles 
6.2 Problem 9: In the figure
given below, PQRS. Show that _{}QPO + _{}ORS = _{}POR
Construction: Draw a line TU parallel to PQ through
O, extend SR to Y, extend QP to X, extend RO to V and extend OP to Z.
Soluition :
No 
Statement 
Reason 

1 
_{}TOP= _{}XPZ 
Corresponding
angles(XQTU) 

2 
_{}XPZ=_{}QPO 
Vertically opposite angle 

3 
_{}QPO= _{}TOP 
Step 1 and 2 

4 
_{}ROT = _{}VOU 
Vertically opposite angle 

5 
_{}VOU = 180^{0 }_{}TOV 
TOU is straight line 

6 
_{}TOV = _{}YRV 
Corresponding angles(TUYS) 

7 
_{}VOU = 180^{0 }_{}YRV 
Step 5 and 6 

8 
180^{0 }_{}YRV = _{}ORS 
YS is straight line 

9 
_{}VOU =_{}ORS 
Step 7,8 

10 
_{}ROT = _{}ORS 
Step 4,Step 9 

11 
_{}POR = _{}POT +_{}TOR = _{}QPO+_{}ORS 
Step 3 ,Step 10 
6.2 Summary of learning
No 
Points learnt 
1 
Axioms,
Postulates and Enunciations on lines 
Additional points:
1. Two lines
which are parallel to the same line are parallel to each other.
2. Two lines
which are perpendicular to the same line are parallel to each other.
3. Two
parallel lines make equal intercepts on all transversals perpendicular to them.
4. ‘Equal intercepts property’: If three or more
parallel lines make equal intercepts on one transversal, then they make equal
intercepts on
any other transversal as well.
5. ‘Proportional intercepts property’: Three or more
parallel lines intersecting any two transversals make intercepts on them in the
same proportion.
Dividing line segment in a given ratio:
We use the ‘Proportional intercepts property’
for this construction.
6.2 Problem 11: Divide a line of 5cm in the ratio of 2:3:2
No 
Construction 

1 
Draw a line AB = 5cm 

2 
Draw a line AC such that AC is
different from AB 

3 
Cut the line AC into 7(=2+3+2) equal line segments of any length
starting from A (by arcs of equal length) 

4 
Join the last marked point(C_{7})
on AC with B to get the line C_{7}B 

5 
Draw parallel lines to C_{7}B,
from C_{2 }and C_{5 }to the line AB to cut the line AB at P and
Q respectively 

6 
Join C_{2}P and C_{5}Q 

7 
AP:PQ:QB = 2:3:2 
Note: If we draw lines parallel to C_{7}B
from C_{1},C_{2},C_{3},C_{4},C_{5},C_{6}
to AB (dotted lines in the figure), these lines cut AB into 7 equal parts
(‘Equal intercepts property’)