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10M0601Similar And Congruent Figures

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The two figures shown above are of the same boy. What can you say about these pictures? Are they similaror congruent?

They are similar but not congruent. How?

Two figures are said to be(i) similar, if they are of the same shape(ii) congruent, if they are of the same shape and size

For example, three equilateral triangles are shown in the given figure.

These three triangles are of the same shape; hence, they are similar. However, the last two triangles are of the samesize and shape. Therefore, the last two triangles are similar and congruent. Thus, we can say that similar figurescan also be congruent if they are of the same size.

Similarly, the three circles shown in the following figure are similar as they are of the same shape. However, only thefirst two circles are congruent as they have the same radii.

1 cm 1 cm 1.5 cm

From these examples, we can conclude that congruent figures are always similar but similar figures neednot be congruent.

Two polygons of the same number of sides are similar if the(i) corresponding angles of the polygons are equal(ii) corresponding sides of the polygons are in the same ratio

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However, if the polygons fulfil only one of the above mentioned criteria, then they are not necessarily similar. Let us tryand understand this with the help of examples:

Consider the following figures:

2.5 cm2.5 cm

2.5 cm

2.5 cm 2.5 cm

2.5 cm

2 cm2 cm

Two quadrilaterals are shown here. The first one is a rectangle and the second is a square. Each internal angle of a squaremeasures 90°. Each internal angle of a rectangle also measures 90°. Thus, the corresponding angles of these quadrilateralsare equal but the corresponding sides are not in the same ratio. Therefore, the given quadrilaterals are not similar.

Let us now try and apply what we have just learnt in some examples.

Example 1:Identify the pairs of similar and congruent figures from the following figures.

(i) 3 cm

2 cm

4 cm

2.5 cm

70º

80º

100º 110º

(ii) 1 cm

(iii) 1.5 cm

1 cm

2 cm

1.25 cm

70º

80º

100º 110º

(iv)

2 cm

2 cm

2 cm 2 cm

(v)

2 cm

2 cm

2.1 cm 2.1 cm (vi) 3 cm 3 cm

3 cm

(vii) 2.5 cm

3 cm

3 cm (viii) 1 cm

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Solution:Figures (i) and (iii) are similar because their corresponding angles are equal and their corresponding sides are in thesame ratio. However, these figures are not congruent as they are of different sizes.

Figures (ii) and (viii) are congruent as they are of the same shape and size (circles with radius 1 cm each).

Example 2:Are the following figures similar or congruent?

Solution:The two given figures show two one-rupee coins. As both the figures represent the same coin in two different sizes,they are similar to each other. However, the pictures are not congruent because of their different sizes.

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10M0602Proof Of Basic Proportionality Theorem

Suppose ABC is any triangle. DE is a line parallel to BC and intersecting the sides AB and AC at D and Erespectively.

A

B

D E

C

Is there any relation between the lengths of the sides AD, DB, AE and EC?

Yes. The relation between the sides of the triangle isAD AEDB EC

.

In other words, we can say that if we have a line parallel to one side of a triangle, then that line divides the other twosides in the same ratio.

This is one of the important theorems of triangles given by Thales. It is also known as Basic proportionality theorem.It can be stated as follows:

“If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, thenthe other two sides are divided by this line in the same ratio”.

We can prove this theorem as follows:

Firstly, let us join BE and CD and then draw DM and EN perpendicular to AC and AB respectively.A

B C

D

N M

E

Now, area ( ADE) = 1

AD EN2

… (1)

Similarly, area ( BDE) = 1

DB EN2

… (2)

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777

Also, area ( ADE) =1

AE DM2

… (3)

area ( DEC) = 1

EC DM2

… (4)

On dividing equation (1) by (2), we obtain:

1AD ENarea ( ADE) AD2 ... (5)

1area ( BDE) DBDB EN2

And on dividing equation (3) by (4), we obtain:

1AE DMarea ( ADE) AE2 ... (6)

1area ( DEC) ECEC DM2

Observe that BDE and DEC are on the same base DE and between the same parallels BC and DE.

Therefore, area ( BDE) = area ( DEC) … (7)

From equations (5), (6) and (7), we obtain:

AD AE

DB ECThus, the line DE divides the side AB and AC in the same ratio.

Now, what can we say about the converse of this theorem? Is the converse also true?Yes, the converse of the basic proportionality theorem is also true and it can be stated as follows:

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

We can prove the converse as follows:

Let a line l intersect the sides AB and AC of ABC at D and E respectively such that AD AE

DB EC.

We have to prove that DE BC.

Let us assume that DE is not parallel to BC. Then, there must be another line parallel to BC. Let DF BC.

A

B C

D E

F

l

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By Basic proportionality theorem, we have:

AD AF...(1)

DB FCBut it is given that:

AD AE...(2)

DB ECFrom (1) and (2), we obtain:

AF AE

FC ECOn adding 1 on both the sides, we obtain:

AF AE1 1

FC ECAF+FC AE+EC

FC ECAC AC

FC ECFC = EC

This is possible only when points F and E coincide with each other, i.e. DF is the line l itself.But DF BC

l BC or DE BC

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

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