6.4 Triangles:
6.4.1 Classification and Theorems on triangles
The
word Triangle can be split as tri and angle, meaning three angles. Definition: ‘Triangle’
is a closed figure formed by three distinct (noncollinear) line segments. Note
that triangle has Three Sides : AB, BC, CA Three Vertices : A, B, C Three Angles : _{}ABC, _{}BCA, _{}CAB 

Depending upon the measure of angles and sides,
triangles are classified as:
Classification 
Type 
Property 
Example 
Based on
angles 
Acute angled
triangle 
Each angle is less
than 90^{0} 

Right angled
triangle 
One angle is 90^{0} _{}PQR =90^{0} 


Obtuse angled
triangle 
One angle is more
than 90^{0} _{}BCA > 90^{0} 


Equiangular
triangle 
All angles are equal _{}PQR =_{}QRP =_{}RPQ = 60^{0} 


Isosceles triangle 
Two angles are equal _{}ABC =_{}BCA ^{} 


Based on
Sides 
Scalene triangle 
All sides are of
different length AB≠BC≠CA 

Equilateral
triangle 
All sides are equal AB=BC=CA 


Isosceles triangle 
Two sides are equal PQ=PR 


Isosceles right
angled triangle 
A right angled
triangle with 2 sides equal _{}ABC =90^{0} And AB=BC 

We know that sum of angles in a triangle is 180^{0}.
Let us prove this, mathematically.
6.4.1 Theorem 1: In any triangle
sum of the three angles is 180^{0}.
Data: ABC is a triangle
To prove: _{}ABC+_{}BAC +_{}ACB = 180^{0}
Construction: Draw a line DE parallel to BC and
passing through A
No 
Statement 
Reason 

1 
_{}EAB = _{}ABC 
By theorem on parallel lines as
EF is  BC and AB is transversal. 

2 
_{}EAC = _{}ACB 
By theorem on parallel lines as
EF is  BC and A is transversal. 

3 
_{}EAB+_{}BAC +_{}EAC= 180^{0} 
Postulate 4: sum of angles on
the straight line = 180^{0} 

4 
_{}ABC+_{}BAC +_{}ACB= 180^{0} 
Substitute ABC for EAB, ACB for
EAC in step3 
6.4.1 Problem 1: If one angle of an isosceles triangle is 40^{0}
find the other two angles.
Solution:
Note: In an isosceles triangle two
angles are equal. We also know that sum of all angles in a triangle is 180^{0}.
We have two scenarios:
1.
Let x be the common angle and 40^{0 }be the other angle, then we have
x
+ x + 40^{0 }= 180^{0 } 2x
=180^{0 } 40^{0 } =
140^{0} _{} x = 70^{0 } The
angles of the triangle are 70^{0},70^{0} and 40^{0} 

2. Let the common angle be 40^{0 }and other
angle be x Then
40^{0 }+ 40^{0 }+ x = 180^{0 } 80^{0
}+ x = 180^{0} _{} x = 100^{0 } The
angles of the triangle are 40^{0},40^{0} and 100^{0}. 

6.4.1 Problem 2: Find the angles of an equiangular (equilateral)
triangle.
Solution:
Note:
All The angles of an equiangular (equilateral) triangle are equal. Also sum
of all angles in a triangle = 180^{0}. If
x is one of the angles of the triangle, then we have x + x + x = 180^{0} i.e. 3x =180^{0}. Therefore x = 60^{0} 

Definition: If any side of a triangle is extended, the angle formed
at the vertex is called an ‘exterior’
angle. In
the figure _{}ACD is Exterior Angle The
two angles inside the triangle, opposite to the adjacent angleof the exterior angle are called ‘interior opposite’ angles. In
the figure _{}BAC and _{}ABC are Interior
opposite Angles 

Figure 
Exterior Angle 
Interior opposite Angles 
k 
_{}ACD Observe here that exterior angle is an obtuse angle i.e. _{}ACD > 90^{0}) 
_{}BAC and _{}ABC 

_{}CBD Observe here that exterior angle is an acute angle i.e. _{}CBD < 90^{0}) 
_{}BAC and _{}ACB 

_{}ABD Observe
here that exterior angle is a right angle i.e. _{}ABD =90^{0}) 
_{}BAC and _{}BCA 
Note that since triangles have three sides,
they have three exterior angles.
6.4.1 Theorem 2: If one of the sides of a triangle is extended, the
exterior angle so formed is equal to sum of interior opposite angles.
Data: ABC is a triangle
To Prove: _{}ACD = _{}ABC + _{}BAC
Proof:
No 
Statement 
Reason 

1 
_{}ABC+_{}BCA +_{}CAB = 180^{0} 
Theorem
: Sum of the angles in a triangle = 180^{0} 

2 
_{}BCA+_{}ACD = 180^{0} 
Postulate
4: sum of angles on the straight line = 180^{0} 

3 
_{}ABC+_{}BCA +_{}CAB = _{}BCA+_{}ACD 
Axiom
1 

4 
_{}ABC + _{}BAC= _{}ACD 
Axiom
2 ( The equals added here is _{}BCA) 
Observations:
No 
Corollaries of above two
theorems(6.4.1) 
Reasoning( if x,y,z are three
angles of a triangle) 
1 
Exterior
angle > each of the interior opposite angles 
If
x,y > 0 then x+y >x and x+y > y 
2 
A
triangle can not have more than one right angle 
If
x+y+z =180, both x and y can not be 90 
3 
A
triangle can not have more than one obtuse angle 
If
x>90 then y+z <90 
4 
In
every triangle at least two angles are acute 
With
x < 90, Both y, z can not be >90 
5 
In
a right angled triangle the sum of two
other angles = 90^{0} 
If
x=90 then y+z has to be 90 and hence both
y,z<90 
6 
If
two angles in one triangle are equal to two angles in another triangle then
third angle of both the triangles are equal 
x+y+z
= 180 and hence z =180xy 
6.4.1 Problem 3: An exterior
angle of a triangle is 90^{0} and one interior opposite angle is 45^{0}.
Find the remaining angle of the triangle.
Solution:
Let
x be the other interior angle of the triangle. We
know that, exterior angle of a triangle = sum of interior opposite angles. _{}90^{0} = x+45^{0} Hence
x = 45^{0}. 

6.4.1 Problem 4: Find all the angles in the triangle given
below.
Solution:
Since
100^{0} is the exterior angle at vertex B, its interior opposite
angles are p and q. _{}p+q = 100^{0 } (1) Since
130^{0} is the exterior angle at vertex C, its interior opposite
angles are r and q. _{}r+q = 130^{0 }(2) We
also know that p + q + r = 180^{0} 100^{0}
+ r = 180^{0} (By substituting value of p+q in the above equation) _{} r = 180^{0}100^{0}=80^{0} By substituting this value of r in (2) we get q= 50^{0}.^{} By substituting this value of q in (1) we get p= 50^{0}.^{} Therefore
50^{0},50^{0} and 80^{0 }are the values of p, q and r
respectively. 

6.4.1 Problem 5: Prove that sum of the
angles of a quadrilateral is 4 right angles.
Solution:
Any
quadrilateral can be divided in to 2 triangles and sum of angles in a
triangle is 2 right angles. _{}Sum of angles in a quadrilateral = 2*(sum of angles in each
of the triangle) = 2*180^{0} =360^{0} 

6.4.1 Problem 6: If in a triangle ABC, 2(_{}A20) = _{}B+10= 2(_{}C10), find each of the angle
Solution:
Note
A+B+C =180 and hence B = 180CA Since
it is given that 2(A20) = _{}B+10 I.e. 2A40
=B+10 I.e.
2A = B+50 =
(180CA)+50 = 230
–C –A. By
transposition we get 3A = 230C
(1) Since
it is given that 2(A20) = 2(C10) I.e.
A20 = C10 I.e. A = C+10 (2) Substituting
this value of A in (1) we get 3A
= 3C+ 30 = 230C I.e. 4C = 200 (On transposition of –C
and 30) _{} C = 50 On
substituting this value of C in (2) we get A =60. Substituting
value of A and C in A+B+C = 180 we get B = 70 _{} The angles are A =
60, B=70 and C=50. 

6.4.1 Problem 7: Prove that bisectors of any two adjacent
angles of a rhombus form a right angled triangle
We have to prove that POQ
= 90^{0}.
Proof:
No 
Statement 
Reason 

1 
_{}SPQ+_{}PQR =180^{0} 
PQRS
is a rhombus, and hence PSQR and interior angles are supplementary 

2 
2(_{}OPQ+_{}PQO) = 180^{0} 
It
is given that 

3 
_{}OPQ+_{}PQO = 90^{0} 
Step
2 

4 
_{}POQ = 180 (_{}OPQ+_{}PQO) = 90^{0} 
Sum
of angles in a triangle = 180^{0} 
6.4.2 Construction of
Triangles:
We have seen that a triangle has 3 measurable
angles and 3 measurable sides and thus in all has six elements.
But to construct a triangle uniquely, we do not
need all the six elements. We just need three elements, one of which has to be
a side.
6.4.2.1. Construction
of a triangle when 3 sides are given
6.4.2 Problem 1: Construct a triangle ABC with AB = 3cm, BC = 4cm
and AC = 5cm.
Method:
First draw a rough figure of the triangle ABC.
No 
Steps 

1 
Draw a line and mark a point A
on that line 

2 
With A as center, cut an arc of
radius 3cm to the cut the above line at B (AB=3cm) 

3 
With A as center cut a large arc
of radius 5cm. 

4 
With B as center cut a large arc
of radius 4cm. 

5 
Let the arcs drawn in steps 3
and 4 meet at C 

6 
Join AC and BC: ABC is the
required triangle 
6.4.2 Problem 2: A field is in the shape of an equilateral
triangle. Perimeter of the field is 2490 Meters. Construct the triangle with
suitable scale.
Solution:
We
know that the perimeter of a triangle is the sum of all its side. Since it is
given that the field is an equilateral triangle, all its sides are equal. Therefore
3*sides = 2490 Meters: side = 2490/3 = 830Meters Since
it will be difficult to construct a triangle of side = 830 meters, we can use
the scale of 100meters = 1cm to construct the triangle. We need to construct
an equilateral triangle with side = 8.3cm. Since
all the sides of an equilateral triangle are same, we need to construct a
triangle with sides 8.3cm, 8.3cm and 8.3cm. Exercise: Follow the method described in
6.4.2 problem 1(above) to construct the triangle. 

6.4.2.2. Construction
of a triangle when 2 sides and an included angle are given
6.4.2 Problem 3: Construct a triangle ABC with AB = 3cm, BC = 4cm
and _{}ABC =120^{0}
Method:
First draw a rough figure of the triangle ABC
No 
Steps 

1 
Draw a line and mark a point A,
on that line 

2 
With A as center, cut an arc of
radius 3cm to cut the above line at B(AB=3cm) 

3 
Use protractor to mark a point
whose angle is 120^{0} from B 

4 
Draw a line from B passing
through the above point 

5 
With B as center, cut an arc of
radius 4cm to cut the above line at C 

6 
Join AC : ABC is the
required triangle 
Note: The above method
is used for constructing a right angled triangle with 2 sides given.
6.4.3 Congruency of Triangles
You must have seen ponds. Have you ever thought of finding
the width of the pond without getting into the water?
Similarly will it be possible to find the width of
a river without getting into water?
Geometry helps us in solving such problems which we
encounter in our daily life.
If you look in a dictionary, we find that
congruence means equivalence, resemblance, similarity, etc. In geometry we say
that,
two figures are congruent if they have similar
properties (i.e. when we super impose one figure over the other, they fit in
exactly without variations).


2
Lines
are
congruent if they are of equal lengths 
2 Angles are congruent if they are equal measure in degrees 
The
‘corresponding sides’ are the sides opposite
to the angle which are equal in two triangles (In
the above figure, AC and DF,AB and EF, BC and DE are corresponding sides) The
‘corresponding angles’ are the angles opposite
to the sides which are equal in two triangles. (In
the above figure, _{}ABC and _{}DEF, _{}ACB and _{}EDF, _{}BAC and _{}DFE are corresponding angles) Two Triangles are said to be ‘congruent’ if the corresponding three sides and three angles of both triangles are equal. 
AB=CD=3cms 
_{}ABC =_{}FDE = 60^{0} 
AC=DF,AB=EF, BC=ED, _{}CAB =_{}EFD_{}ABC=_{}DEF,_{}ACB=_{}EDF 
The symbol for
congruence is_{} 

AB_{}CD 
_{}ABC_{}_{}FDE 
_{}ABC_{}_{}DEF 
Observation: Two congruent triangles have equal area because they fit
in exactly when superimposed.
6.4.3 Example 1: Though a triangle has 6 elements (3 sides and 3
angles) let us construct the below mentioned triangles (ABC)
with the following three elements (data).
1. BC = 4cm,
CA = 4.5cm, BA= 5cm
2. BC = 4cm, _{}ABC =40^{0}, _{}BCA =50^{0}
3. BC = 5cm, CA=7cm, _{}BCA = 35^{0}
4. The angles of the triangle are 60^{0}, 50^{0}, 70^{0}
^{ }
Note that when we are given only three angles, we
can construct several triangles.
(In Figure 4 and Figure 5, although the corresponding
angles are same, AB _{} DE, BC _{} FE and AC _{} DF)
Conclusion: With the following three elements we can construct unique triangles
1. Length of three
sides
2. Two angles and
length of one side
3. Length of two sides
and the included angle
Note: Given, length of two sides and an
angle other their included angle we can not construct a unique triangle.
In the adjacent figure,
length AB and AC are given and also the angle _{}BAC.
Extend AC and let D be
a point on this extended line such that BC =BD.
We observe that with
the given data, we have drawn two triangles ABC and ABD (AB as common side
and _{}BAC as common angle).
Table A: In general we can tabulate our
findings as follows:
Side 
Side 
Side 
Angle 
Angle 
Angle 
Result 
Y 
Y 
Y 
 
 
 
We
can construct triangle uniquely 
Y 
 
 
Y 
Y 
 
We
can construct triangle uniquely 
Y 
Y 
 
Y 
 
 
We
can construct triangle uniquely (if the angle is included angle) 
 
 
 
Y 
Y 
Y 
Several
triangles can be constructed 
Note: If 2 angles are given, the third angle can easily be found as the sum of 3 angles in a triangle is 180^{0}.^{}
Observation: As per the definition
of congruency, for congruence, all the corresponding 3 sides and 3 angles of
two triangles have to be equal.
But we have seen that, to draw a unique triangle we
do not need to know all the 6 elements (length of all the 3 sides and all the 3
angles).
We have concluded that we need a maximum of three
elements. We have also found out these combinations (Table A).
We have seen that it is possible to construct a
triangle, if length of two sides and the included angle are given. We conclude
that
‘Two triangles are
congruent if two sides and included angle of one triangle are equal to the corresponding
sides and included angle of the
other triangle’.
This statement is called ‘SAS (Side, Angle, and Side) Postulate on congruence’.
6.4.3 Problem 1: To Measure width of a pond
Find the width of the pond given in the figure.
Solution:
Step:
Have 2 poles (at A, B) on the edges of the pond, where you want to find the
width. Have
another pole (at C) on the ground in front of the pond, which is visible to
both A and B. Extend
AC to E such that AC=CE and BC to D such that BC=CD. _{}ACB = _{}DCE (vertically opposite Angles) By
the above SAS postulate we conclude that the Triangles ABC and DEC are
congruent. Hence
AB=DE Measure
distance DE to know the width of the pond. 

6.4.3 Problem 2: In the adjoining figure PQRS is square. M is the
middle point of PQ.
Prove that SM=RM
Solution:
Since
PQRS is square, PS=QR, _{}SPQ =90^{0 } & _{}PQR = 90^{0 }(so _{}SPQ = _{}PQR ) Since
M is the middle point of PQ, PM=MQ. So
we have two triangles SPM and MQR whose 2 sides and the included angles are
equal. Therefore
SPM _{} MQR and hence their
3^{rd} sides (SM, MR) are equal. 

Activity: Construct a triangle
of AB=4cm and AC=BC=5cm (2 sides are equal). Measure
the angles _{}CAB and _{}ABC. Do
you notice that _{}CAB =_{}ABC? 

Observations: In a triangle, angles
opposite to equal sides are equal.
Let us prove this mathematically.
6.4.3 Base Angle Theorem:
The angles opposite to equal sides of a triangle
are equal.
Data: In the adjoining triangle ABC, AC=BC
To prove _{}CAB= _{}ABC
Construction: Bisect the angle _{}ACB such that the bisector line CD meets AB at
point D.
Steps 
Statement

Reason 

1 
AC=BC 
This
is the given data (equal sides) 

2 
_{}ACD=_{}DCB 
CD
is bisector of the angle _{}ACB 

3 
CD
is the common side of triangle ACD and DCB 
our
construction 

4

Triangles
ACD & DCB are congruent 
SAS
Postulate (2 sides and included angles are equal) 

5 
_{}CAB= _{}ABC 
Corresponding
angles of congruent triangles 
This proves the theorem.
6.4.3 Problem 3: In the adjoining figure AB=AC. L and M are points
on AB and AC such that
Also prove that _{}ABM _{}_{}ACL and _{}LCB _{}_{}MBC, LMBC
Solution:
Steps 
Statement

Reason 

1 

given 

2 
BL
= CM 
AB=AC
and AL=AM(given) 

3 
_{}ALM =_{}LMA 
The
angles opposite to equal sides(AL,AM) of a triangle are equal (Base Angle
theorem) 

4 
AB=AC 
Given. 

5 
_{}BAM is common to both _{}ABM and _{}ACL 


6 
_{}ABM _{} _{}ACL 
SAS
postulate (step1,step5,step4) 

7 
_{}ABC =_{}BCA 
Base
angle theorem(AB=AC) 

8 
LB=CM 
Step1,2,3 

9 
BC
is common base to both _{}LCB and _{}MBC 


10 
_{}LCB _{}_{}MBC 
SAS
postulate (step2,step7,step9) 

11 
2_{}ALM = 180^{0}_{}LAM 
_{}ALM+_{}LMA+_{}LAM = 180^{0} and _{}ALM = _{}LMA 

12 
2_{}ABC = 180^{0}_{}LAM 
_{}ABC+_{}BCA+_{}LAM = 180^{0} and _{}ABC = _{}BCA 

13 
_{}ALM = _{}ABC 
Step
11 and 12 as RHS of both are same 

14 
LM
BC 
Corresponding
angles are equal (Step 13) 
Activity: Construct few pairs of
triangles such that, two angles (say 30^{0},50^{0}) on the
common side of one triangle are equal to
other triangles. Did you notice that the lengths of
opposite sides are equal in all the cases?
6.4.3 Converse of Base Angle Theorem: In a triangle, the sides which
are opposite to equal angles are equal.
Data: In the adjoining triangle ABC, _{}CAB= _{}ABC
To prove:
AC=BC
Construction: Bisect the angle _{}ACB such that the bisector line CD meets AB at
point D.
Steps 
Statement

Reason 

1 
_{}CAB= _{}ABC 
This
is the given data (equal sides) 

2 
CD
is the common side of triangle ACD and DCB 
our
construction 

3 
_{}ACD =_{}DCB 
Construction
(bisector of _{}ACB) 

4

ACD
_{} DCB 
ASA
Postulate 

5 
AC=BC 
Corresponding
sides of congruent triangles 
This proves the Converse of the base
angle theorem.
6.4.3 Problem 4: Prove that in an isosceles triangle, the angular
bisector of vertex angle bisects the base and is perpendicular to the base.
Solution:
Data: In the adjoining triangle ABC, AC=BC
To prove: AD=DB and _{}ADC =_{}CDB = 90^{0}
Construction: Bisect the angle _{}ACB so that the bisector line CD meets AB at point
D.
Steps 
Statement

Reason 

1 
AC=BC 
This
is the given data (equal sides) 

2 
_{}ACD = _{}BCD 
our
construction 

3 
CD
is the common side of triangle ACD and DCB 
our
construction 

4 
Triangles
ACD & DCB are congruent 
SAS
Postulate (2 sides and included angles are equal) 

5 
AD=DB 
In
a congruent triangle corresponding sides are equal 

6 
_{}ADC =_{} CDB 
In
a congruent triangle corresponding angles are equal 

7 
_{}ADC+_{}CDB=180^{0} 
Two
angles are on a straight line 

8 
_{}ADC =_{}CDB = 90^{0} 

Activity: Construct few pairs of
triangles such that the sides of a triangle (say 4cm,5cm,6cm) are equal to the sides
of other triangles.
Did you notice that they are all congruent?
We have seen that it is possible to construct a
triangle if lengths of all the 3 sides are given. We conclude that
‘Two triangles are
congruent if the sides of one triangle are equal to the corresponding sides of
another triangle’.
This statement is called ‘SSS (Side, Side, Side) Postulate
on congruence’.
6.4.3 Problem 5:
PQRS is a square. A, B, C, D are mid points of PQ, QR, RS and SP
respectively. Prove that _{}BAC=_{}BCA.
Solution:
Since
PQ=SR and A and C are mid points of PQ and SR, we have AQ=CR. Since
B is mid point of QR, we have QB=BR. Since
PQRS is square _{}AQB=90^{0} =_{}BRC So
we have 2 triangles, AQB and CRB, whose 2 sides are equal and included angles
are also equal. By
SAS postulate they are congruent and hence AB=BC and CAB is an isosceles
triangle. Thus
by base angle theorem, angles opposite to equal sides are equal. Hence
_{}BAC=_{}BCA 

Activity: Construct few pairs of triangles such that 2 angles and
their common side (say 60^{0},70^{0
}, 4cm^{ }) are equal to 2 angles and
the common side of other triangles. Did you notice
that they are all congruent?
We have seen that it is
possible to construct a triangle if two angles and length of their common side
are given. We conclude that
‘Two triangles are congruent if two angles and common side of
one triangle are equal to the corresponding angles and common side of another
triangle’.
This statement is
called ‘ASA
(Angle, Side, Angle) Postulate on congruence’.
Corollary: (corollary means
consequence, outcome…) “Two
triangles are congruent if two angles and any one side of one triangle are
equal
to
corresponding 2 angles and corresponding side of another triangle”. This statement is
called ‘AAS
(Angle, Angle, Side ) Condition
on congruence’.
Proof:
1. Sum of
all the angles in a triangle =180^{0}
2. Given 2
angles of a triangle, we can arrive at the third angle of triangle (=180^{0}
– sum of given 2 angles)
3. Any 2 of
the 3 angles can become the angles on the given common side.
Since two triangles have corresponding angles equal
and common sides equal, by ASA postulate, these triangles are congruent.
6.4.3 Converse of Base Angle Theorem: In a triangle, the sides which
are opposite to equal angles are equal.
Data: In the adjoining triangle ABC, _{}CAB= _{}ABC
To prove:
AC=BC
Construction: Bisect the angle _{}ACB such that the bisector line CD meets AB at
point D.
6.4.3 Problem 6: To Measure width of river
Solution:
Identify
a fixed object such as tree (B) on the other side of river. Erect a pole at A
(opposite to B) on
your side, such that BA is a straight line. Erect a pole at some distance
from A at C. Erect
another pole at D such that AC=CD (C is mid point of AD). Erect another pole
at E such
that DE is perpendicular to AD and points B,C and E lie on a straight line. We
notice the following: 1. _{}BAC = 90^{0}= _{}CDE (BA and DE are constructed perpendicular to
AD). 2.
_{}ACB = _{}DCE (Vertically opposite angles) 3.
AC=CD (By construction) Thus
2 angles and their common side in _{}BAC and _{}EDC are equal. By
ASA postulate,_{}BAC _{}_{}EDC. Thus BA=DE. By
measuring DE we get the width of river without getting into water. 

6.4.3 Problem 7: In the adjoining figure AC Bisects DF and _{}EDC =_{}AFE. Prove that AE=EC
Solution:
In
the given figure_{}AEF =_{}DEC (opposite angle) DE=EF
& _{}EDC =_{}AFE (given data) Therefore
by ASA postulate, _{}AEF and _{}DEC are congruent triangles. _{}AE=EC. 

Table: Postulates for congruency of triangles:
Side 
Side 
Side 
Angle 
Angle 
Angle 
Postulate 
Y 
Y 
Y 
 
 
 
SSS 
Y 
 
 
Y 
Y 
 
ASA 
Y 
Y 
 
Y 
 
 
SAS 
Note that for congruency of triangles, at least one side has to be equal.
6.4.3 Exercise : Use these postulates to prove that your method of
construction of following is
correct(Refer 6.1)
1.
Construction of an angular bisector
2.
Construction of a perpendicular line at a point on a line(Perpendicular
bisector Theorem)
3.
Construction of a perpendicular bisector of a line
6.4.3 Theorem: Two right angled triangles are congruent if the hypotenuse
and a side of one triangle are equal to the hypotenuse and the
corresponding side of the other triangle.
Data: _{}ABC and _{}DEF are given right angled triangles (_{}ABC =_{}DEF= 90^{0}) and AB=DE, AC=DF
To prove: _{}ABC _{} _{}DEF
Construction: produce FE to the point G such that
GE=BC and join DG.
Steps 
Statement 
Reason 

1 
AB=DE 
Given data 

2 
_{}ABC =_{}DEF= 90^{0} 
Given data 

3 
_{}ABC =_{}DEG = 90^{0} 
Construction and _{}DEG+_{}DEF = 180^{0} 

4 
BC=GE 
construction 

5 
_{}ABC _{}_{}DEG 
SAS postulate(step1,step3,step4) 

6 
_{}ACB=_{}DGE 
Corresponding angles 

7 
DG=AC 
Corresponding sides 

8 
AC=DF 
Given data 

9 
DG=DF 
step7,step8 

10 
DE is common to _{}DEF and _{}DEG 
Construction 

11 
_{}DEG = _{}DEF = 90^{0} 
Construction 

12 
_{}DFE = _{}DGE 
Base angle theorem for_{}GDF 

13 
_{}GDE = _{}EDF 
Sum of 3 angles in a triangle
=180^{0} _{}GDE =180^{0 }–_{}DEG_{}DGE =180^{0 }–_{}DEF _{}DFE(step2,step3,step12) = _{}EDF 

14 
_{}DEG _{}_{}DEF^{} 
ASA (step 13,step10,step11) 

15 
_{}ABC _{}_{}DEF 
Step 5,14 
This is also called RHS (Right Angle, Hypotenuse, Side)
Postulate on congruence.
6.4.3 Problem 8: If the three altitudes of a triangle are equal, prove that it is an
equilateral triangle.
Solution:
The
perpendicular drawn from a vertex of a triangle to its opposite side is called
altitude.
In the adjoining figure
EC,BF,AD are altitudes
Steps 
Statement

Reason 

1 
Consider the _{}BEC and _{}BFC 

2 
EC=BF 
Equal
altitudes(given) 

3 
_{} BEC=_{}BFC = 90^{0} 
BE
and BF are Altitudes 

4 
BC
is common 


5 
_{}BEC _{}_{}BFC 
RHS
postulate 

6 
_{}ABC = _{}BCA 
Corresponding
angles 

7 
Consider the _{}ADB and _{}ADC 

8 
_{} ADB=_{}ADC = 90^{0} 
AD
is Altitude 

9 
AD
is common 


10 
_{}ABC = _{}BCA 
Step 6 

11 
_{}ADB _{}_{}ADC 
ASA
postulate 

12 
AB
=AC 
Corresponding
sides are equal 

13 
BC=
AC 
Similarly
we can prove _{}BFC _{}_{}BFA 

14 
AB=AC=BC 
Step
12,13 
6.4 Summary of learning
No 
Points to remember 
1 
In
any triangle sum of the three angles is 180^{0} 
2 
If
one of the sides of a triangle is extended, the exterior angle so formed is
equal to the sum of interior opposite angles 
3 
Two
Triangles are congruent if the corresponding three sides and three angles of
both triangles are equal 
4 
SAS
Postulate 
5 
The
angles opposite to equal sides of a triangle are equal (Base angle theorem)
and converse of this is also true 
6 
SSS
Postulate 
7 
ASA
Postulate 
Additional Points:
More constructions of triangles:
6.4.2 3. Construction
of a triangle when two angles and included side(base) are given
6.4.2 Problem 4: Construct a triangle ABC with _{}ABC = 40^{0}, _{}BCA = 50^{0} and BC = 3cm
1.
First draw a rough figure of the triangle ABC. 2.
Draw the line BC=3cm 3.
At B and C draw lines at 40^{0} and 50^{0} to base BC and let
these lines meet at A 4.
ABC is the required triangle 

6.4.2 4. Construction
of a right angled triangle when lengths of one side and hypotenuse are given
6.4.2 Problem 5: Construct a right angled triangle with base of
length 3cm and hypotenuse of length 5cm.
1.
First draw a rough figure of the triangle ABC. 2.
Draw the line BC=3cm 3.
Draw a perpendicular at B 4.
From C draw an arc of radius 5cm to cut the perpendicular at A 5.
ABC is the required triangle 

6.4.2 5. Construction
of an isosceles triangle when its base and altitude (height) are given
6.4.2 Problem 6: Construct an isosceles triangle ABC with base AB =
6cm and altitude = 4cm.
Here,
we use the property of an isosceles triangle which states that the
altitude bisects the base. 1.
First draw a rough figure of the triangle ABC. 2.
Draw the base AB = 6cm 3.
Bisect AB at D (AD=DB=3cm) 3.
Draw a perpendicular at D above AB 4.
From D draw an arc of radius 4cm to cut this perpendicular at C 5.
ABC is the required triangle 

6.4.2 6. Construction
of an isosceles triangle when its altitude and angle of vertex are given
6.4.2 Problem 7: Construct an isosceles triangle with altitude =
4.5cm and angle at vertex = 50^{0}
Here,
we use the property of an isosceles triangle which states that, the altitude
bisects the angle at vertex. 1.
First draw a rough figure of the triangle ABC. 2.
Draw a base line 3.
Chose any point D on this line 4.
Draw perpendicular at D above the base line 4.
From D draw an arc of radius 4.5cm to cut this perpendicular at C 5.
Draw lines making an angle of 25^{0} with CD, on both sides of DC and
let these lines cut the base line at A and B 6.
ABC is the required triangle 

6.4.2 7. Construction
of an equilateral triangle when its altitude is given
6.4.2 Problem 8: Construct an equilateral triangle whose altitude =
4.5cm
Here, we use the property of an equilateral
triangle which states that, each angle
in an equilateral triangle is 60^{0}.
This construction is equivalent to the
construction 6.4.2.6 with angle at vertex = 60^{0}.
6.4.2.8. Construction
of an isosceles triangle when its base and base angle is given
(Since base angles of an isosceles
triangle are equal, this construction is equivalent to the construction as
given in 6.4.2.3)
6.4.2.9. Construction
of an equilateral triangle when its one side is given
(Since all sides of an equilateral
triangle are equal, this construction is equivalent to the construction when
all the 3 sides of a triangle are given as in 6.4.2.1)
6.4.3 Perpendicular bisector Theorem:
1. Every point on the perpendicular bisector is
equidistant from the two given points.
2. Any point which is equidistant from the two
given points, lies on the perpendicular bisector of the line joining these 2
points.
In the adjacent figure YLX is the perpendicular bisector
of AB (i.e.
We are required to prove that,
1.
If point P is on YLX then AP=BP
2.
If P is a point such that AP=BP then P lies on the line
XLY
3.
Hint: Part
1: If
point P is L itself then it is obvious that
AP=PB If
point is not L then join AP and PB Use
SAS postulate to prove that _{}ALP _{}_{}BLP (hence AP=BP) This
proves the first part of the theorem. Part
2: If
point P is L itself then P lies on XLY If
point P is not L then join AP and PB Use
SSS postulate to prove that _{}ALP _{}_{}BLP (hence _{}ALY =_{}BLY = 90^{0}) _{} LP is perpendicular
to AB and hence P is on the perpendicular line XLY This
proves the second part of the theorem. 

Note:
The locus of a point which is
equidistant from two fixed points is the perpendicular bisector of the line
segment joining the fixed points.
(Refer section 6.1 for definition of locus).
6.4.3 Angular bisector Theorem:
1. Every point on the anglular bisector is
equidistant from the sides of the angles.
2. Any point which is equidistant from the sides of
the angles lies on the bisector of the angle.
In the adjacent figure AR is the bisector of _{}BAC (i.e. _{}EAR = _{}DAR)
We are required to prove that,
1.
If point P is on AR then PE=PD
2.
If P is a point on AR such that PE = PD then P lies on the
line AR
Hint: Draw
PE and PD perpendicular to AC and AB respectively from P Part
1: Use
ASA postulate to prove that _{}APE _{}_{}APD (hence EP=DP) This
proves the first part of the theorem Part
2: Use
RHS postulate to prove that _{}APE _{}_{}APD (hence _{}EAP =_{}DAP) This
proves the second part of the theorem. 

Note:
The locus of a point which is
equidistant from two intersecting straight lines is the angular bisector of the
angle made by these lines.
6.4.1 Observations:
1. The ratio of the areas of two triangles is equal
to the ratio of the products of the base and it’s corresponding height, of
those two triangles.
Proof:
Let
ABC and DEF be the two triangles as shown in the adjoining figure. We
know that the area of a triangle = 1/2*Base*height _{}Area of ABC/Area of DEF =
(1/2) BC* =
BC* Based
on the above observation, prove the following: 2.
Triangles of equal heights have areas proportional to their corresponding
bases. 3.
Triangles of equal bases have areas proportional to their corresponding
heights. 

Inequalities: (Proof not provided for the following theorems)
Theorem 1: If two sides of a triangle are not equal, then the angle
opposite to the greater side is greater than the angle opposite
to the smaller side.
In
the adjoining figure _{}BCA > _{}ABC 

Theorem 2: If two angles of a triangle are not equal, then the side
opposite to the greater angle is greater than the side opposite to the smaller angle.
In the adjoining figure AB > BC
Corollaries:
1. The sum
of lengths of any two sides of a triangle is always greater than the third side
(Ex. AC+BC>AB)
2. The
difference between the lengths of any two sides of a triangle is always less than
the third side
(Ex. ABAC
< BC)
3. Among
all the line segments joining a point outside a given line and any point on the
line, the
perpendicular
line segment is the shortest.