2. ACKNOWLEDGEMENT
We wish to express our deep gratitude and sincere thanks to the
Principle Sir, Dr. S.V. Sharma for his encouragement and for all the
facilities that he provided for this Project. We sincerely appreciate this
magnanimity by taking us into his fold for which we shall remain
indebted to him. We extend our hearty thanks to Mrs. Shweta Bhati ,
Maths teacher, who guided us to the successful completion of this
Project. We take this opportunity to express our deep sense of gratitude
for her invaluable guidance, constant encouragement, immense
motivation, which has sustained our effort at all the stages of this
project work…
We can`t forget to offer our sincere thanks to parents and also to all
group members who helped us to carry out this activity successful and
for their valuable advice and support, which we received from them
time to time……..

3. This is to certify that this project on
“TRIANGLES (SIMILARITY)” is made by
Group – 4 and the members are Abhishek
Mahto, Lakshya Kumar, Mohan Kumar, Ritik
Kumar, Vivek Singh of class X E . We are
thankful to Principal sir ‘Dr. S.V. Sharma’
for giving us this opportunity and
concerned Subject teacher ‘Shweta Bhati’.Principal Sign. Teacher Sign. Marks Obtained
VIDYA BAL BHAWAN SR. SEC. SCHOOL

4. Content
CONCEPT OF SIMILARITY
SIMILAR POLYGONS
SIMILAR TRIANGLES AND THEIR
PROPERTIES
SOME BASIC RESULTS ON
PROPORTIONALITY
THALE`S THEOREM
CONVERSE OF BPT
CRITERIA FOR SIMILARITY OF
TRIANGLES
AREAS OF SIMILAR TRIANGLES

5. INTRODUCTION
In general, there are several objects which have something common between them.
Observing them closely, let see that some of them have same shape but may have
different or same size. For example, if the photographs a person developed from
same negative, they all look same in all respect except for their size. Such objects are
called similar objects. Two line segments of different sizes, two circles of different
radii, two squares of different sizes, two rectangles of different dimensions - come
under similar figures. One smaller circle can be got by shrinking a larger circle. One
bigger square can be got by stretching a smaller square. Then, what about the
similarity of triangles? Is it true to say any two given triangles are similar? The
answer is NO. This is true only when the triangles are equilateral.

7. In earlier classes, we have learnt about
congruent figures.
Two geometric figures having the same shape
and size are known as congruent figures. Note
that congruent figures are alike in every
respect.
Geometric figures having the same shape but
different sizes are known as similar figures.
Two congruent figures are always similar but
similar figure not be congruent.

8.  ILLUSTRATION 1 : Any two line segments are always
similar but they need not be congruent. They are congruent, if
their lengths are equal.
 ILLUSTRATION 2 : Any two circles are similar but not
necessarily congruent. They are congruent if their radii are
equal.

9.  ILLUSTRATION 3 :
(i) Any two squares are similar.
(ii) Any two equilateral triangles are similar.

10. SIMILAR POLYGONS
DEFINITION : Two polygons are said to be
similar to each other, if
(i) Their corresponding angles are equal
(ii) The length of their corresponding

11. If two polygons ABCDE and PQRST are similar,
then :
Angle at A = Angle at P, Angle at B = Angle at Q,
Angle at C = Angle at R, Angle at D = Angle at S,
Angle at E = Angle at T
And, AB BC CD DE EA
PQ QR RS ST TP
IF two polygons ABCDE and PQRST, are similar,
we write ABCDE ̴ PQRST

12. Here, ABCDE ̴ PQRST
D
A B QP
S
130⁰
100⁰
120⁰
90⁰
130⁰
100⁰
120⁰

13. SIMILAR TRIANGLES AND
THEIR PROPERTIES
DEFINATION : Two triangles are said to be
similar, if their
(i) Corresponding angles are equal
(ii) Corresponding sides are proportional
It follows from this defination that two triangles
ABC and DEF are similar, if

14. (i) A= D, B= E, C= F and,
(ii) AB BC AC
DE EF DF
CB
A
FE
D

15. NOTE 1: In the later part of this PPT we
shall show that two condition given in
previous slide are not independent. In
fact, if either of the two conditions
hold, then the oyher holds
automatically. So any one of the two
conditions can be used to define
similar triangles.
NOTE 2: If corresponding angles of two
triangles are equal, then they are
known as equiangular triangles

16. SOME BASIC RESULTS ON
PROPORTIONALITY
In this section, we shall discuss some basic results
on proportionality.
Let us do the following activity.
ACTIVITY: Draw any angle XAY and mark points
P1,P2,D,P3 and B on its arm Ax such that
AP1=P1P2=P2D=DP3B=1 units.
Through point B, draw any line intersecting arm AY at point
C. Also, through point , draw a line parallel to BC to
intersect AC at E.

17. We have,
AD=AP1+P1P2+P2D=3 units
And, DB=DP3+P3B=2 units
AD 3
DB 2
Now, measure Ae and EC and find AE
EC
You will find that
AE 3
EC 2
AD AE
DB EC
Thus, we observe that in ABC if De II BC, then
AD AE
DB EC
We prove this result as a theorem known as basic proportionality theorem or Thale`s
Theorem.
C
xBP3DP2P1
A
E
Y

18. THALE`S THEOREM
(Basic Proportionality Theorem or Thale`s
Theorem) If a line is drawn parallel to one side
of a triangle intersecting the other two sides,
then it divides the two sides in the same ratio

19. Given : A triangle ABC in which DE II BC,
and intersects Ab at D AC in E.
To prove : AD AE
DB EC
Construction: Join BE,CD and
draw EM I BA and DN I CA
Proof:
Area of triangle
Similarly,
Hence,

20. Similarly,
Triangles BDE and DEC are on the same base, i.e. DE and
between same parallels, i.e. DE and BC.
Hence,
From above equations, it is clear that;

21. PRESS ENTER AFTER VIDEO PLAYBACK FOR NEXT SLIDE

22. If a line divides any two sides of a triangle in
the same ratio, then the line is parallel to the
third side.

23. Construction: ABC is a triangle in which line DE divides AB and AC in the
same ratio. This means:
To prove: DE || BC
Let us assume that DE is not parallel to BC.
Let us draw another line DE' which
is parallel to BC.
Proof:
If DE' || BC, then we have;
According to the theorem;
Then according to the first theorem; E and E' must be coincident.
This proves: DE || BC

24. PRESS ENTER AFTER VIDEO PLAYBACK FOR NEXT SLIDE

25. CRITERIA FOR SIMILARITY OF
TRIANGLES
AAA or AA (ANGLE- ANGLE)
SSS (SIDE- SIDE- SIDE)
SAS (SIDE-ANGLE-SIDE)

26. AA CRITERIA
If in two triangles, corresponding angles are
equal, then their corresponding sides are in the
same ratio (or proportion) and hence the two
triangles are similar. This is also called AAA
(Angle-Angle-Angle) criterion.

27. Construction: Two triangles ABC and DEF are drawn so that
their corresponding angles are equal.
This means:
∠ A =∠ D, ∠ B = ∠ E and ∠ C = ∠ F
To prove:
Draw a line PQ in the second triangle so that DP = AB and PQ = AC
Proof:
Because corresponding sides of these two triangles are equal
This means; ∠ B = ∠ P = ∠ E and PQ || EF

28. This means;
Hence;
Hence,

29. SSS CRITERIA
If in two triangles, sides of one triangle are
proportional to the sides of the other triangle,
then their corresponding angles are equal and
hence the two triangles are similar. This is also
called SSS (Side-Side-Side) criterion.

30. Construction: Two triangles ABC and DEF are drawn so that
their corresponding sides are proportional.
This means:
To prove: ∠ A = ∠ D, ∠ B = ∠ E
and ∠ C = ∠ F
And hence; ABC ~  DEF
In triangle DEF, draw a line PQ so that DP = AB and DQ = AC
Proof:
Because corresponding sides of these two triangles are equal
This means;

31. This also means; ∠ P = ∠ E and ∠ Q = ∠ F
We have taken; ∠ A = ∠ D, ∠ B = ∠ P and ∠ C = ∠ Q
Hence; ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F
From AAA criterion;
 ABC ~ DEF proved

32. SAS CRITERIA
If one angle of a triangle is equal to one angle
of the other triangle and the sides including
these angles are proportional, then the two
triangles are similar. This is also called SAS
(Side-Angle-Side) criterion.

33. Construction: Two triangles ABC and DEF are drawn so that one of the angles of
one triangle is equal to one of the angles of another triangle. Moreover, two
sides included in that angle of one triangle are proportional to two sides
included in that angle of another triangle. This means;
∠ A = ∠ D and
To prove: ABC ~ DEF
Draw PQ in triangle DEF so that,
AB = DP and AC = DF
Proof:
Because corresponding sides of these two triangles are equal
(GIVEN)
∠ A = ∠ D

34. Hence;
( from SSS criterion)
Hence;
Hence;ABC ~DEF proved

36. AREAS OF SIMILAR TRIANGLES
THEOREM: The ratio of areas of two similar
triangles is equal to the square of the ratio of
their corresponding sides.

37. Construction: Two triangles ABC and PQR are drawn so that,
ABC ~ PQR.
To prove:
Draw AD ⊥ BC and PM ⊥ PR
Proof:
Hence;

38. Now, in ABD and PQM;
∠ A = ∠ P, ∠ B = ∠ Q and ∠ D = ∠ M (because ABC ~ PQR)
Hence; ABD ~ PQM
Hence;
Since, ABC ~ PQR
So,
Hence;
Similarly, following can be proven:

40. THEOREM: If a perpendicular is drawn from the
vertex of the right angle of a right triangle to the
hypotenuse then triangles on both sides of the
perpendicular are similar to the whole triangle and
to each other.

41. Construction: Triangle ABC is drawn which is right-angled at B. From
vertex B, perpendicular BD is drawn on hypotenuse AC. To prove:
ABC ~ ADB ~ BDC
Proof:
In ABC and ADB;
∠ ABC = ∠ ADB
∠ BAC = ∠ DAB
∠ ACB = ∠ DBA
From AAA criterion; ABC ~ ADB
In ABC and BDC;
∠ ABC = ∠ BDC
∠ BAC = ∠ DBC
∠ ACB = ∠ DBC
From AAA criterion; ABC ~ BDC
Hence; ABC ~ ADB ~ BDC proved.

42. Pythagoras Theorem
In a right triangle, the square of the
hypotenuse is equal to the sum of the
squares of the other two sides.

43. Construction: Triangle ABC is drawn which is right angled at B. From
vertex B, perpendicular BD is drawn on hypotenuse AC. To prove:
Proof:
In ABC and ADB;
Because these are similar triangles (as per previous theorem)
In ABC and BDC;
Adding equations (1) and (2), we get;
Proved.

45. PRESENTATION BY
NAME ROLL NO.
ABHISHEK MAHTO 1
LAKSHYA KUMAR 13
MOHAN KUMAR 15
RITIK KUMAR 23
VIVEK SINGH 38