Triangle notes summarized.

Chap ter




4
1.0 INTRODUCTION

2.0 CRITERIA FOR CONGRUENCE OF TRIANGLES
2.1 Side-Angle-Side (SAS) congruence criterion
2.2 Angle-Side-Angle (ASA) congruence criterion
2.3 Angle-Angle-Side (AAS) congruence criterion
2.4 Side-Side-Side (SSS) congruence criterion
2.5 Right Angle-Hypotenuse-Side (RHS) congruence criterion

3.0 PROPERTIES OF AN ISOSCELES TRIANGLE

4.0 INEQUALITIES IN A TRIANGLE

EXERCISE-1
(ELEMENTARY)
EXERCISE-2
(SEASONED)
EXERCISE-3 (CBSE- (SA-I) PATTERN
) EXERCISE-4 (COMPETITIVE
ASSESSMENT)
UnitOne

, Triangles

TRIANGLES (TRI)
(TRI – 1)
1.0 INTRODUCTION
Congruent figures : Two geometrical figures, having exactly the same shape and size
are known as congruent fig. For congruence, we use the symbol 
Thus, two line segments are congruent, if they have the same
length. Two angles are congruent, if they have the same
measure.
Congruent triangles : Two triangles are
A D
congruent, if and only if one of them can be made
to superimpose on the other, so as to cover it
exactly.
Thus, congruent triangles are exactly identical, i.e.,
their corresponding three sides and the three B C E
F
angles are equal.
If ABC is congruent to DEF, we write ABC  DEF. This happens when AB =
DE, BC = EF, AC = DF and A = D, B = E, C = F.
In this case, we say that the sides corresponding to AB, BC and AC are DE, EF and DF
respectively, and the angles corresponding to A, B and C are D, E and F
respectively.
Thus, the corresponding parts of two congruent triangles are equal. We show it by the
abbreviation C.P.C.T., which means corresponding parts of congruent triangles.
Congruence relation in the set of all triangles : From the definition of
congruence of two triangles, we obtain the following results :
(i) Every triangle is congruent to itself i.e. ABC  ABC.
(ii) If ABC  DEF, then DEF  ABC.
(iii) If ABC  DEF, and DEF  PQR, then ABC  PQR.

2.0 CRITERIA FOR CONGRUENCE OF TRIANGLES
In earlier classes, we have learnt some criteria for congruence of triangles. Here, in
this class we will learn the truth of these either experimentally or by deductive
proof.
In the previous section, we have studied that two triangles are congruent if and only if
there exists a correspondence between their vertices such that the corresponding sides and
the corresponding angles of two triangles are equal i.e. six equalities hold good, three of
the corresponding sides and three of the corresponding angles.
In this section, we shall prove that if three properly chosen conditions out of the six
conditions are satisfied, then the other three are automatically satisfied. Let us now
discuss those three conditions which ensure the congruence of two angles.
2.1 Side-Angle-Side (SAS) congruence criterion
s
l
e Axiom : Two triangles are congruent if two sides and included angle of one triangle are
n
g equal to
a
the two sides and the included angle of the other triangle.
A D
ir
.T
In the given figure (A) & (B) in ABC and DEF,
4
we have :
\
I
X
\
AB = DE, AC = DF and A = D.
cs
 ABC  DEF [By SAS- Remark : SAS Congruence
t
i criteria] rule holds
m
a
e
t
9

,B C E F
but ASS or SSA rule does not hold. (A) (B)




9

, Class IX : Mathematics
2.2 Angle-Side-Angle (ASA) congruence criterion
Theorem : Two triangles are congruent, if two angles and the included side of one
triangle are equal to two angles and the included side of the other triangle.
Given : Two triangles ABC and DEF (fig. A & B) such that
B = E, C = F and BC = EF.
To prove : ABC  DEF..
Proof : To prove ABC  DEF, we need to consider three possible situations :
A D




B C E F
(A) (B)
Case 1. Let AB = DE.
Statement Reason
1. In ABC and DEF
(i) AB = DE Supposed
(ii)B= E Given
(iii)BC = EF Given
2. ABC  DEF By SAS criteria
Case 2. Let AB > DE (Fig. A & B).
A D
H


B C E F
(A (B)
)
Construction : Take a point H on AB such that HB = DE.
Statement Reason
1. In HBC and DEF
(i) HB = DE By
Construction
(ii) B = E Given
(iii) BC = EF Given
2. HBC  DEF By SAS criteria
3. HCB  DFE C.P.C.T.
4. ACB  DFE Given
5. HCB  ACB From (3) and
(4)
It can be possible only if H coincides with A. In other words, AB = DE.
Case 3. Let AB < DE (Fig. A & B).
A D s
l
e
n
g

M a

ir
.T


B C E F
4
\
I
(A (B) X
\
cs
)
t
i

Construction : Take a point M on DE such that ME = AB. By repeating the same m
a
e
t

9