Class 10

### Topic to be covered

☼ Similarity of Triangles.

### Similarity of Triangles

● You may recall that triangle is also a polygon. So, we can state the same conditions for the similarity of two triangles. That is:
Two triangles are similiar, if
(i)color{red}{" their corresponding angles are equal"}
(ii) color{red}{"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.

● To understand the Basic Proportionality Theorem, let us perform the following activity:

=> Activity 2 : Draw any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and
B such that color{red}{AP = PQ = QD = DR = RB.}

Now, through B, draw any line intersecting arm AY at C (see Fig. 6.9).

● Also, through the point D, draw a line parallel to BC to intersect AC at E. Do you observe from your constructions that (AD)/(DB) = 3/2  ? Measure A E and EC. What about (AE)/(EC) ? Observe that (AE)/(EC) is also equal to 3/2. Thus, you can see that in Δ ABC, DE || BC and (AD)/(DB) = (AE)/(EC) Is it a coincidence? No, it is due to the following theorem (known as the Basic Proportionality Theorem):

=>color{blue}{"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 (see Fig. 6.10).

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 Δ ADE ( =1/2 text(base × height)) = 1/2 AD xx EN

Recall from Class IX, that area of Δ ADE is denoted as ar(ADE).

So, ar(ADE) = 1/2 AD xx EN

Similarly, ar(BDE) =1/2 DB xx EN

ar(ADE) = 1/2 AE xx DM and ar(DEC) = 1/2 EC xx DM

● Therefore {ar(ADE)}/{ar(BDE)} = (1/2 AD xx EN)/(1/2 DB xx EN) = (AD)/(DB) .........(1)

and {ar(ADE)}/{ar(DEC)} = (1/2 AE xx DM)/(1/2 EC xx DM) = (AE)/(EC) .................(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 :

color{blue}{(AD)/(DB) = (AE)/(EC)}

"Activity 3 :" Draw an angle XAY on your notebook and on ray AX, mark points B_1, B_2,B_3, B_4 and B such that AB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B.

Similarly, on ray AY, mark points
C_1, C_2, C_3, C_4 and C such that AC_1 = C_1C_2 = C_2C_3 = C_3C_4 = C_4C. Then join B_1C_1 and BC (see Fig. 6.11).

● Note that (AB_1)/(B_1 B) = (AC_1)/(C_1 C) (Each equal to 1/4)

You can also see that lines B_1C_1 and BC are parallel to each other, i.e.,

B_1C_1 || BC ...................(1)

(AB_2)/(B_2 B) = (AC_2)/(C_2 C) = (2/3) and B_2C_2 || BC .............(2)

(AB_3)/(B_3 B) = (AC_3)/(C_3 C) = (3/2) and B_3C_3 || BC ...............(3)

(AB_4)/(B_4 B) = (A C_4)/(C_4 C) = (4/1) and B_4C_4 || BC ...............(4)

● From (1), (2), (3) and (4), it can be observed that if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

● You can repeat this activity by drawing any angle XAY of different measure and taking any number of equal parts on arms AX and AY . Each time, you will arrive at the same result. Thus, we obtain the following theorem, which is the converse of theorem 6.1:

●=>color{blue}{"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 (see Fig. 6.12).

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

So (AD)/(DB) = (AE')/(E'C)

Therefore color{red}{(AE)/(EC) = (AE')/(E'C)}

● Adding 1 to both sides of above, you can see that E and E′ must coincide.
Q 3139380212

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)

Class 10 Chapter 6 Example 1
Solution:

DE ∥ BC (given)

So (AD)/(DB) = (AE)/(EC)

or (DB)/(AD) = (EC)/(AE)

or (DB)/(AD) +1 = (EC)/(AE) +1

or (AB)/(AD) = (AC)/(AE)

so (AD)/(AB) = (AE)/(AC)
Q 3149380213

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)
Class 10 Chapter 6 Example 2
Solution:

Let us join AC to intersect EF at G (see Fig. 6.15).

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)

(AG)/(GC) = (BF)/(FC) .....................(2)

Therefore, from (1) and (2),

(AE)/(ED) = (BF)/(FC)
Q 3179380216

In Fig. 6.16, (PS)/(SQ) = (PT)/(TR) and ∠ PST = ∠ PRQ. Prove that PQR is an isosceles triangle.
Class 10 Chapter 6 Example 3
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.