Triangles

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

  1. 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

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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.

3.png

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.

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6.3 Similarity of Triangles

  1. 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.

10.png

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,

12.png—————-(1)

 

and

13—————-(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.

14.png
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:

17

 

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 .

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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)

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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.

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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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