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`☼` 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.

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

Class 10 Chapter 6 Example 1

`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

Class 10 Chapter 6 Example 2

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

Class 10 Chapter 6 Example 3

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.