1. INTRODUCTION
Congruent figure:

2. Congruent figure:

3. Congruent figure:

4. A B C D E F
Congruent triangles: Two triangles are said to be congruent if they have the same size and same shape.
Similar Figures: Two figures having the same shape (and not necessarily the same size) are called similar figures. B A C R Q P

5. SIMILAR FIGURES
Here two more circles do not have same radius, they are not congruent to each other but all of them have same shape. So they all are Similar.
What about two (or more) squares or two (or more) equilateral triangles. As observed in case of circles here also all squares and equilateral triangles are similar.
From these figures we can say that all congruent figures are similar but the similar figures are not congruent.
(i)
(ii)
(iii)

6. P S A D Q R B C
Two polygons of same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
Scale Factor: Same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygon.

7. Basic Proportionality Theorem
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. A N M D B E C
We are given a  ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.
We need to prove that
Let us join BE and CD and then draw DM  AC and EN  AB.
Now ,

8. Since we know that area of  ADE is denoted as ar(ADE)
Note that BDE and DEC are on same base and between same parallel lines BC and DE.
So,
Similarly,
Therefore,
and A N M D B E C

9. Converse of BPT
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side.
This theorem can be proved by taking a line DE such that and assuming that DE is
not parallel to BC.
If DE is not parallel to BC, draw a line DE’ parallel to BC. A D E’ E B C
(BPT)
Adding 1 on both sides of above, we can see that E and E’ coincides.

10. Criteria for Similarity of Triangles
In  ABC and  DEF, if
A= B, B= E, C= F and
, then the two triangles are
similar A B C D E F

11. Theorem (A-A-A Similarity)
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. A B C
This criteria is referred to as the AAA (Angle-Angle-Angle) criteria of similarity of two triangles.
Given: Two triangles ABC and DEF such that A = D, B = E and C = F.
To Prove:  ABC   DEF D E F P Q

12. Cut DP = AB and DQ = AC and join PQ.
So,  ABC   DPQ (SAS Congruency)
This gives B = P = E and PQ||EF
Therefore, A B C
(BPT) D E F P Q
i.e.,
(given)
Similarly,
Hence,  ABC   DEF

13. Theorem (S-S-S Similarity)
If in two triangles, sides of one triangle are proportional to (i.e. in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.
This criteria is referred to as the SSS (Side-Side-Side) similarity criterion for two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that (<1)
Cut DP = AB and DQ = AC and join PQ. A B C
It can be seen that D E F P Q
Therefore,
So,
Thus,  ABC   DPQ
So, A=D, B=E and C=F
Hence,  ABC   DEF

14. Theorem (S-A-S Similarity)
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 criteria is referred to as the SAS (Side-Angle-Side) similarity criterion for two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that
Cut DP = AB and DQ = AC and join PQ.
Now, PQ || EF and  ABC   DPQ
So, A =D, B =P and C =Q
Therefore,  ABC   DEF A B C D E F P Q

15. Theorem
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. A M B C
We are given two triangles ABC and PQR such that  ABC   PQR
We need to prove that
For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles. P Q N R
Now,
and
So,
(1)

16. (AA Similarity criterion)
Now, in  ABM and  PQN,
(As  ABC   PQR)
and
(Each is of 90o)
So,
(AA Similarity criterion) A M B C
Therefore,
(2)
(Given)
Also,
So,
(3)
Therefore,
[from (1) and (3)] P Q N R
[from (2)]
Now using (3), we get

17. Theorem
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
We are given right triangle ABC right angled at B.
We need to prove that AC2 = AB2 + BC2
Let us draw BD  AC
Now,  ADB   ABC
AD . AC = AB (1)
Also,  BDC   ABC
Or, CD . AC = BC (2)
Adding (1) and (2),
AD . AC + CD . AC = AB2 + BC2
AC(AD + CD) = AB2 + BC2
AC . AC = AB2 + BC2
AC2 = AB2 + BC2 B D A C

18. Theorem
In a right triangle, if square of one side is equal to sum of the squares of the other two sides, then the angle opposite the first side is a right angle. A
Here, we are given a triangle ABC in which AB2 + BC2 = AC2
We need to prove that B = 90o.
To start with, we construct a  PQR right angled at Q such that PQ = AB and QR = BC
Now, from  PQR, we have:
PR2 = PQ2 + QR (Pythagoras theorem)
Or, PR2 = AB2 + BC (by construction) (1)
But AC2 = BC2 + AB (given) (2)
So, AC = PR [from (1) & (2)] (3)
Now, in  ABC and  PQR,
AB = PQ (by construction)
BC = QR (by construction)
AC = PR (proved above)
So,  ABC   PQR (SSS congruency)
 B = Q (CPCT)
But Q = 90o (by construction)
So, B = 90o C B Q P R