`color{red} ♦` Angle Sum Property of a Triangle

In the earlier classes, you have studied through activities that the sum of all the angles of a triangle is `180°`. We can prove this statement using the axioms and theorems related to parallel lines.

`color {blue} text(Theorem 6.7) :` The sum of the angles of a triangle is `180º ` .

`color {blue} text(Proof) :` Let us see what is given in the statement above, that is, the hypothesis and what we need to prove. We are given a triangle `PQR` and `angle 1 , angle 2` and `angle 3` are the angles of `Delta PQR` (see Fig. `6.34`).

We need to prove that ` angle 1 + angle 2 + angle 3 = 180°`. Let us draw a line `XPY` parallel to `QR` through the opposite vertex `P`, as shown in Fig. `6.35`, so that we can use the properties related to parallel lines.

Now, `XPY` is a line.

Therefore, ` angle 4 + angle 1 + angle 5 = 180°` (1)

But `XPY || QR` and `PQ, PR` are transversals.

So, `angle 4 = angle 2` and `angle 5 = angle 3`

(Pairs of alternate angles)

Substituting `angle 4` and `angle 5` in (1), we get

` angle 2 + angle 1 + angle 3 = 180°`

That is, `angle 1 + angle 2 + angle 3 = 180°`

Recall that you have studied about the formation of an exterior angle of a triangle in the earlier classes (see Fig. `6.36`). Side `QR` is produced to point `S, angle PRS` is called an exterior angle of `Delta PQR`.

Is `angle 3 + angle 4 = 180°` ? (Why?) (1)

Also, see that

`angle 1 + angle 2 + angle 3 = 180°` (Why?) (2)

From (1) and (2), you can see that

`angle 4 = angle 1 + angle 2`.

This result can be stated in the form of a theorem as given below:

`color {blue} text( Theorem 6.8)` :

If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

It is obvious from the above theorem that an exterior angle of a triangle is greater than either of its interior apposite angles.

Now, let us take some examples based on the above theorems.

`color {blue} text(Theorem 6.7) :` The sum of the angles of a triangle is `180º ` .

`color {blue} text(Proof) :` Let us see what is given in the statement above, that is, the hypothesis and what we need to prove. We are given a triangle `PQR` and `angle 1 , angle 2` and `angle 3` are the angles of `Delta PQR` (see Fig. `6.34`).

We need to prove that ` angle 1 + angle 2 + angle 3 = 180°`. Let us draw a line `XPY` parallel to `QR` through the opposite vertex `P`, as shown in Fig. `6.35`, so that we can use the properties related to parallel lines.

Now, `XPY` is a line.

Therefore, ` angle 4 + angle 1 + angle 5 = 180°` (1)

But `XPY || QR` and `PQ, PR` are transversals.

So, `angle 4 = angle 2` and `angle 5 = angle 3`

(Pairs of alternate angles)

Substituting `angle 4` and `angle 5` in (1), we get

` angle 2 + angle 1 + angle 3 = 180°`

That is, `angle 1 + angle 2 + angle 3 = 180°`

Recall that you have studied about the formation of an exterior angle of a triangle in the earlier classes (see Fig. `6.36`). Side `QR` is produced to point `S, angle PRS` is called an exterior angle of `Delta PQR`.

Is `angle 3 + angle 4 = 180°` ? (Why?) (1)

Also, see that

`angle 1 + angle 2 + angle 3 = 180°` (Why?) (2)

From (1) and (2), you can see that

`angle 4 = angle 1 + angle 2`.

This result can be stated in the form of a theorem as given below:

`color {blue} text( Theorem 6.8)` :

If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

It is obvious from the above theorem that an exterior angle of a triangle is greater than either of its interior apposite angles.

Now, let us take some examples based on the above theorems.

Q 3230567412

In Fig. `6.37`, if `QT bot PR, angle TQR = 40°` and `angle SPR = 30°`, find `x` and `y`.

Class 9 Chapter 6 Example 7

Class 9 Chapter 6 Example 7

In `Delta TQR, 90° + 40° + x = 180°`

(Angle sum property of a triangle)

Therefore, `x = 50°`

Now, `y = angle SPR + x` (Theorem 6.8)

Therefore, `y = 30° + 50°`

`= 80°`

Q 3240567413

In Fig. `6.38`, the sides `AB` and `AC` of `Delta ABC` are produced to points `E` and `D` respectively. If bisectors `BO` and `CO` of `angle CBE` and `angle BCD` respectively meet at point `O`, then prove that ` angle BOC = 90^0 - 1/2 angle BAC.`

Class 9 Chapter 6 Example 8

Class 9 Chapter 6 Example 8

Ray `BO` is the bisector of `angle CBE`.

Therefore, `angle CBO = 1/2 angle CBE`

`= 1/2 (180° – y)`

`= 90° – y/2 ` (1)

Similarly, ray `CO` is the bisector of `angle BCD`

Therefore, ` angle BCO = 1/2 angle BCD`

` = 1/2 (180° – z)`

` = 90^0 - z/2` (2)

In `Delta BOC, angle BOC + angle BCO + angle CBO = 180°` (3)

Substituting (1) and (2) in (3), you get

So, ` angle BOC = z/2 + y/2`

or, ` angle BOC = 1/2 (y + z) ` (4)

But , ` x + y + z = 180°` (Angle sum property of a triangle )

Therefore, ` y + z = 180° – x`

Therefore, (4) becomes

`angle BOC = 1/2 (180° – x)`

` = 90^0 - x/2 `

` = 90^0 - 1/2 angle BAC`