Class 9 Angle Sum Property of a Triangle for CBSE-NCERT

### Topic covered

color{red} ♦ Angle Sum Property of a Triangle

### 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.
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
Solution:

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
Solution:

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