From the above, we can say that *all congruent figures are similar but the similar figures need not be congruent.*

*Two polygons of the 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).*

Note that the same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygons.

*2 . Two polygons of the same number of sides are similar, if *

**(i) all the corresponding angles are equal and **

**(ii) all the corresponding sides are in the same ratio (or proportion).**

From the above, you can easily say that below quadrilaterals ABCD and PQRS are similar

**Remark :** You can verify that if one polygon is similar to another polygon and this second polygon is similar to a third polygon, then the first polygon is similar to the third polygon.

In the two quadrilaterals (a square and a rectangle), corresponding angles are equal, but their corresponding sides are not in the same ratio.

So, the two quadrilaterals are not similar.

In the two quadrilaterals (a square and a rhombus), corresponding sides are in the same ratio, but their corresponding angles are not equal. Again, the two polygons (quadrilaterals) are not similar.

**6.3 Similarity of Triangles**

*Two triangles are similiar, if*

*(i) their corresponding angles are equal and*

*(ii) their corresponding sides are in the same ratio (or proportion).*

Note that **if corresponding angles of two ****triangles are equal, then they are known as ****equiangular triangles**. A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows:

*The ratio of any two corresponding sides in two equiangular triangles is always the same.*

It is believed that he had used a result called the **Basic Proportionality Theorem** (now known as the **Thales Theorem)** for the same.

**Theorem 6.1 :** *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.*

**Proof :**

We are given a triangle 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 AD/DB = AE/EC

Let us join BE and CD and then draw DM⊥AC and EN⊥AB

NOW, area of ΔADE = ½ x (base x height) = ½ x AD x EN

We know that, the area of ΔADE is denoted as ar(ADE)

So, ar(ADE) = ½ x AD x EN

Similarly, ar(BDE) = ½ x DB x EN

ar(ADE) = ½ x AE x DM and ar(DEC) = ½ x EC x DM

Therefore,

—————-(1)

and

—————-(2)

Note that Δ BDE and DEC are on the same base DE and between the same parallels

BC and DE.

So, ar(BDE) = ar(DEC) ———————————-(3)

Therefore, from (1), (2) and (3), we have :

AD/DB = AE/EC

**Theorem 6.2 :** If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

This theorem can be proved by taking a line DE such

that

AD/DB = AE/EC and assuming that DE is not parallel to BC.

If DE is not parallel to BC, draw a line DE′ parallel to BC.

So, AD/DB = AE’/E’C

Therefore, AE/EC = AE’/E’C

Adding 1 to both sides of above, you can see that E and E′ must coincide.

Let us take some examples to illustrate the use of the above theorems.

Example 1 : If a line intersects sides AB and AC of a Δ ABC at D and E respectively and is parallel to BC, prove that AD/AB = AE/AC

**Solution:**

Given, DE||BC,

AD/DB = AE/EC (THEOREM 1.6)

So, DB/AD = EC/AE

so, (DB/AD) + 1 = (EC/AE) + 1

AB/AD = AC/AE

AD/AB =AE/AC

**Example 2 :** ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB (see Fig. 6.14). Show that AE/ED = BF/FC .

**Solution:**

Let us join AC to intersect EF at G from figure. AB || DC and EF || AB (Given) So, EF || DC (Lines parallel to the same line are parallel to each other)

Now, in Δ ADC,

EG || DC (As EF || DC)

So, AE/ED = AG/GC (Theorem 6.1) (1)

Similarly, from Δ CAB

CG/AG = CF/BF

i.e., AG/GC = BF/FC

Therefore, from (1) and (2), AE/ED = BF/FC

**Example 3 :** In Fig., PS/SQ = PT/TR and ∠ PST = ∠ PRQ. Prove that PQR is an isosceles triangle.

**Solution :**

It is given that PS/SQ = PT/TR⋅ So, ST || QR (Theorem 6.2) Therefore, ∠ PST = ∠ PQR (Corresponding angles)—————–(1)

Also, it is given that ∠ PST = ∠ PRQ —————–(2)

So, ∠ PRQ = ∠ PQR [From (1) and (2)]

Therefore, PQ = PR (Sides opposite the equal angles) i.e., PQR is an isosceles triangle.

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