1. Geometry - IX Class
Chapter No. 3
CONGRUENCE OF TRIANGLES

2. Congruent Figures
Look at the figures, and tell which of them have same shape and
size
A B C D C D
P Q
R
A
B
C
P
R
R

3. It is a bit difficult to determine it by mere observation.
One may think of using
i) divider ii) a trace paper or
iii) drawing the figures on the plane paper and cut along the
boundary and superpose the pieces.
Such method of Superposition is not a proper way to
determine it.
We say figures of the same shape and size are
congruent figures.
A B
C D
Q
RS
P
A
B C
P
Q R

4. Association with real number and congruency
We associate a unique real number with the figures of the same
type, which helps to determine the congruency.
(i) The real number associated with the segment is its length
and that associated with angle is its measure.
We know l(AB) = l (CD) seg AB ≈ seg CD
and m ABC = m PQR
(ii) To determine the congruency of triangles, we suggest on
activity leading to the concept of one to one correspondence.

5. One to one correspondence and congruency of
triangles.
Activity :
To determine whether the given ∆ABC and ∆ PQR are congruent
......
(i) Draw the given triangles on plane papers.
(ii) Cut the triangular pieces along the boundary.
(iii) Put one triangular piece over the other and try to match.
(iv) See whether the triangles cover each other exactly.
(v) Try to find out all possible ways of keeping one triangular
piece over the other.
(vi) What do you find ?
B
A
C
P
RQ

6. There are six different ways of putting one
triangle cover the other i.e. six different one to
one correspondence between the vertices of
the triangles.
ABC ↔ PQR , ABC ↔ QPR , ABC ↔ QRP
ABC ↔ RQP , ABC ↔ RPQ , ABC ↔ PRQ
Out of six different one to one
correspondences between the vertices of
triangles if there exist at least one, in which
triangles cover each other exactly than we
say the triangles are congruent.

7. Congruent Triangles
Out of the six one to one
correspondence between the vertices
of two triangles, if there exist at-least
one, one to one correspondence such
that, the sides and angles of one triangle
are congruent to corresponding sides
and angles of other triangle then the
triangles are said to be congruent with
respect to that correspondence and the
property known as congruence.

8. We have
ABC ↔ QPR such that
i ) A ≈ Q ii ) B ≈ P iii ) C ≈ R
iv) side AB ≈ side QP
v) side BC ≈ side PR
vi) side AC ≈ side QR
then ∆ABC ≈ ∆ QPR
i.e. To determine the congruency of triangle we require the six
conditions.
But do we really need all the six condition?

9. Sufficient conditions for congruency of the triangles
Three out of six conditions if property chosen are
sufficient to determine the congruency of the two
triangles. When these three conditions are satisfied then
the other three automatically get satisfied, hence the
triangles become congruent. This fact was first proved by
Euclid the father of Geometry.
These sufficient conditions are referred as
SAS, SSS, ASA, SAA tests.
SAS test is taken for granted
i.e. accepted as postulate and other are proved.

10. Activity : Verification of SAS Test :
i) Construct ∆ABC and ∆ PQR of the given measures as shown in the fig.
ii) Cut the triangular pieces along the boundary.
iii) Place triangular piece ABC over triangular piece DEF such that A fall
on D and AB falls along DE.
iv) Since AB = DF , so B falls on E
v) Since A = D , so AC will fall along DF
vi) But AC = DF
C will fall on F Thus AC will coincide with DF and BC will
coincide with EF
∆ ABC coincides with ∆ DEF
Hence ∆ ABC ≈ ∆ DEF
A
B C
P
Q R

11. Included Angle is a must for SAS Test :
Activity :
’
i) Construct another ∆DEF of the same measures
i.e. EF = 5 cm, ED = 3.8 cm, C = 450
ii) See, we get two different triangles of the same measures.
∆ABC ≈ ∆DEF
∆ABC ≈ ∆D‘EF
∆DEF ≈ ∆D’EF
i.e. if the angle is not included then the triangles may or may
not be congruent.
A
B C
5cm
3.8cm
D
E F
5cm
3.8cm D’

12. Think it over
i) ∆ABC is isosceles triangle
with seg AB ≈ seg BC
We know that ∆ABC ≈ ∆ABC ---- (reflexivity)
State another congruent correspondences.
ii) ∆ ABC is equilateral triangle
We have ∆ABC ≈ ∆ABC ---- (reflexivity)
State another congruent correspondences.
iii) ∆ ABC is Scalene triangle
∆ABC ≈ ∆ABC ---- (reflexivity)
State another congruent correspondence if exit.
Also find another congruent correspondences between the
vertices of triangles in each case
i) ∆ABC ≈ ∆PQR and AB = AC (i.e. isosceles)
ii) ∆ABC ≈ ∆PQR and AB = BC = AC (i.e. equilateral)
iii) ∆ABC ≈ ∆PQR (scalene)

13. Isosceles Triangle Theorem and its Converse
In ∆ABC , seg AB ≈ seg BC C = A
Here we suggest activity for the verification
i) Construct ∆ABC where AB = BC
and measure C and A
ii) Construct ∆ABC where C = A
and measure AB and BC
iii) Construct ∆ABC where AB ≠ BC
and measure C and A
iv) Construct ∆ABC where C ≠ A
and measure AB and BC
What relations you get?

14. Proof of the theorem
The proof can be given by two ways
i) ∆ABC ≈ ∆CBD (SAS Test)
A ≈ C
ii) ∆ABC is isosceles triangle and AB = BC
∆ABC ≈ ∆CBA A ≈ C
B
A CD

15. Think it over
We find subtitle “Think it over” on many pages in this chapter. This
will motivate the students to think differently and develop the
higher order thinking skill. (HOTS)
Some instances are given from the text.
E.g. (i) Show : ABP ≈ ACP
This is the solved problem in the text book.
Then it is asked to think it over if the point is in the exterior of ∆ ABC
Show : ABP ≈ ACP
A
B C
P
B C
A
P

16. Think it over
(2) Solved problem
To prove
side AC || side DB
Think it over if
Here
Pair of congruent sides are
changed then think, can
side AC and DB be parallel.
A
B
D
C
O
O
A
B
D
C

17. Think it over
Solved Problem
Show :
Ray AD || Side BC
Think it over if AB ≠ AC
then state whether the ray AD remains parallel to side BC
R
D
B C
A E

18. Think it over
The proof of 300 - 600 - 900 theorem is given
i.e. in 300 - 600 - 900 triangle, length of side
opposite to angle 300 is half the hypotenuse.
i) In turn it is asked to find the length of side
opposite to angle 600
ii) Again it is asked to modify the theorem.
iii) Think it over the converse of 300 - 600 -900
theorem.

19. Think it over
Do we really need two different tests ASA
and SAA, for congruency?
When two angles of triangle are congruent
with corresponding two angles of another
triangle, then the third pair also becomes
congruent.
Hence actually we do not need two different
tests ASA and SAA.

20. Think it over
(1) Medians of isosceles triangle
Solved problem :
Given : AB = AC
BD and CE are medians
Show : BD = CE
Now it is asked to think it over
can medians AE and BD be
equal.
A
B C
DE
A
B C
D
E

21. Think it over
Altitudes of isosceles triangle
Solved example :
Given : Altitude BE = Altitude DC
Show : AB = BC
(1) Asked to think over the converse of above rider
If AB = AC
Show : Altitude DC = Altitude BE
A
B C
ED ┐ ┌
A
B C
ED
F
┐ ┌

22. Now think
If AB = AC then
Can AF = BF
When this is true?
A
B C
ED
F
┐
┐ ┌

23. Think it over
If two triangles are congruent then
corresponding altitudes are equal.
Then think it over the areas of
congruent triangles.
Whether areas of congruent triangles
are equal.

24. Think it over
l1 , l2 , l3 , l4 are different lines intersecting the
seg AB in different points. Which line is
perpendicular bisector of AB?
What are the conditions to be imposed over a line to be
perpendicular bisector of given segment?
P
A B
l1
A B┐Q
l2
A B
R
l3
A B┐S
l4

25. Activity
Leading to perpendicular bisector theorem.
Measure the distances
P1A , P1B
P2A , P2B
P3A , P3B
P4A , P4B
What condition is obeyed by the points P1 , P2 , P3 ------ the points
on the perpendicular bisector of seg AB.
P1
A B
P2
P3
P4
P5

26. Activity
Leading to angle bisector theorem
Measure the distances P1Q1 , P1R1
P2Q2 , P2R2
P3Q1 , P3R1
What condition is obeyed by the points on the angle
bisector of angle?
B
A
C
R1
P2 P3 P4
Q1
Q2
Q3
R4
R3
R2
Q4
P1

27. Activity
Leading to shortest segment theorem
Out of the segments PM, PC, PD, PA, PB
which is shortest and what is the angle made by
shortest segment with line l ?
┐
B A M C D
l
P

28. Locus
i) Points on perpendicular bisector of segment.
ii) Points on angle bisector.
iii) Points on circle.
The set of points which obey certain
conditions is known as locus.

29. Locus
Activity :
AP . is constant
ellipse
AC + BC = constant
When points move according to given conditions, certain path is
traced, that path is known as locus.
P A
A B
C

30. Activity
Leading to
Difference of remaining two sides < Length of a
side of a triangle < Sum of remaining two sides
Take sticks of different sizes and try to
construct triangles. Think about the restrictions on
the lengths of sides of triangles.
Sets of sticks
i)8 cm, 3 cm, 5 cm
ii) 8 cm, 5 cm, 6 cm,
iii) 8 cm, 3 cm, 2 cm.

31. Sum of two sides = Third side Sum of two sides > Third side
Diff. of two sides = Third side Diff. of two sides < Third side
Triangle is possible
Sum of two sides < Third side
Diff. of two sides > Third side
8
5 3
6
5
8
8
3 2

32. Up- gradation of Chapter
In an attempt to upgrade the chapter
i) Proofs of SSS, SAA, ASA tests and converse of
300 - 600 - 900 theorem are included.
ii) Many activities are suggested to verify the
situation.
iii) Day to day life problems, hot problems, historical
problems, application and activity based
problems are included.
iv) At some places the problems based on basic
geometric construction are given, to enrich the
concept of congruence and its application.

33. Day today life Problems
(1) A’ B’
How will you find the distance between two places A and B
when there is a big obstacle between them.
(2) How will you find the
breadth of the river
without crossing it.

34. Activity based Problems
(1)
Two squares of the same size are kept on one
another, as shown in the figure where O is the centre
of one square. Find the area overlapped.
(2) How will you use unmarked ruler with parallel
edges to construct an angle bisector of given angle.
A
B C
D
P
O
R
S

35. Historical Problems
Show
This method of trisection of angle was suggested by Archimedes
CAB
3
1
P
CAB
3
1
P
P
A
B C

36. Application based Problems
(1)
Show : Ray PQ ┴ line l
(2)
Show : Ray BM is angle bisector of
ABC
A P B
C D
Q
l
R
P
B
Q S
M
A
C

37. Higher Order Thinking Skill (HOTS) Problems
(1)
Given : AB < AC
Ray AP is angle
bisector of BAC
Show : BP < PC
(2)
In ∆ABD
seg AD ┴ seg BC , BD < DC
Show : AB < AC
A
B
P
C
A
B
D
C

38. Routine Problems For 1 mark each
1) ∆ABC ≈ ∆PQR , A = 400 , Find P
2) State the triangle
Congruent to ∆ABP
3) In ∆ABC, seg AB ≈ seg BC , B = 400 find A
4) Name the greatest side of ∆ABC
5) ∆ABC ≈ ∆PQR and seg AB ≈ seg BC . State another
correspondence between the vertices of ∆ ABC and ∆ PQR
which is congruent.
6) PQ is shortest segment then
find PQR.
P
B
Q
C
A
P
T Q R S
A
B C
400 300

39. For 2 marks
1) Arrange the sides of
triangle in ascending order of
their lengths.
2) ∆ABD ≈ ∆CDB
To prove this some
information is missing.
State that minimum
information required
to prove it.
A
B C
400 300
A
B C
D

40. 3)
PA = PB
QA = QB
RA = RB
SA = SB
TA = TB
Name the path of points i.e. the locus of points P,Q,R,S............
4) In 300 - 600 - 900 theorem, side opposite to angle 300 is 5 cm,
find other two sides.
5) In ∆ ABC , AB = 5 cm , BC = 7 cm.
then find the maximum possible length of third side of ∆ ABC.
6) With side 4, 3, 1 cm, state whether a triangle can be drawn,
explain.
7) Perpendicular bisector of side AB and BC of ∆ ABC intersect
each other at point P.
If PA = 5 cm, find PB and PC.
. P
. Q
. R
. S
. T
A B

41. For 3 marks
1)
Given : AB = AC
line PQ || side BC
Show : seg AP ≈ seg AQ.
2) Prove that corresponding altitudes of congruent triangles
are congruent.
3)
seg CP ┴ seg AB ,
segBQ ┴ seg AC
seg AB ≈ seg AC
Show : seg PC ≈ seg BQ.
A
P
B C
Q
A
B C
QP ┐ ┌

42. 4)
Given : AB = AC
and B - C - D
Show : AB < AD
5) Given : BP and CP
are angle bisector of
B and C respectively.
Show : Ray AP is angle
bisector of A.
A
B
C
D
A
B

43. For 4 marks
1)
Given : Ray AD is angle
bisector of A and is
perpendicular on
opposite side BC
Show : AB = AC
2)
B ≈ C
and arcs are drawn
with centers B and C
keeping the radius constant.
Show : seg PS || seg BC.
B
A CD
┐┌
A
P
B
Q R
S
C

44. 3)
ABC = 900
D mid point of AC
Show : AD =DC = BD
4) Show that perpendicular bisector of angles of triangle are
concurrent.
5)
Arcs drawn with centre B,
intersect the sides of ABC in
points P, Q and R,S.
Segments PQ and RS
intersect at M.
Show that Ray BM is angle
bisector of ABC.
A
B C
D
R
P
B
Q S
M
A
C

45. For 5 marks
1) The length of any side of a triangle is greater that the
difference between the lengths of remaining sides.
2) Perimeter of a triangle is greater than the sum of the three
medians of triangle.
3)
Show : BAC < BPC < BQC
A
B C
P
Q
R

46. 4)
Fig. ABCDE is a regular
pentagon.
If BP = CQ = DR = ES = AT
then show that fig. PQRST is
a regular pentagon.
5)
ABCD is a square
P, Q, R and S are the points
on its sides
Such that AP = BQ = CR = DS
and ASP = 300
Show that (i) PQRS is quare
(ii) If perimeter of PQRS is
16 cm then find perimeter
ABCD
A
A
B
C
D
E
S
T
P Q
R
D
S
A
P
B
Q
CR

47. Non-Routine Problems
(1) P is a point inside the equilateral ∆ABC
such that PA = 3 cm, PB = 4 cm, PC = 5 cm.
Find the side of ∆ ABC.
(2)
Find the maximum value of C, so that the figure can be
drawn as per the description.
(3) Point P is in the
interior of rectangle ABCD.
Find another point Q in its
interior such that line PQ will
divide the rectangle in to
two regions of equal areas.
A
B C
P Q
D
A B
C
. P

48. Fallacy
A right angle triangle is equilateral triangle
Proof :
Construction : (i) Draw angle bisector of C
(ii) Draw perpendicular bisector of side AB
(iii) Name their point of intersection as M
(iv) Draw seg ML ┴ side AC
seg MN ┴ side CB
(v) Draw MA, MB.
BA
C
L
N
R
M

49. Proof : Point M is on angle bisector of C
-------- (Construction)
LM = NM ------- (i)
Point M is on perpendicular bisector of seg AB
-------- (Construction)
AM = MB ------- (ii)
∆LMA ≈ ∆NMB ------- (side hyp. theorem)
LA = NB ----------- (c.s.c.t.) ------- (iii)
Similarly
∆LMC ≈ ∆NMC
LC = NC ----------- (c.s.c.t.) ------- (iv)
Adding (iii) and (iv)
LA + LC = NB + NC
AC = CB
Hence proved
Find the fallacy in the above proof.
BA
C
L
N
R
M