NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.

In the earlier classes, we 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.

Theorem 6.7

The sum of the angles of a triangle is 180°.

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 ∠1, ∠2 and ∠3 are the angles of triangle PQR (see Figure 6.34).

Figure 6.34

 

We need to prove that ∠1 + ∠2 + ∠3 = 180°.

Let us draw a line XPY parallel to QR through the opposite vertex P, as shown in Figure 6.35, so that we can use the properties related to parallel lines.

Now, XPY is a line. Therefore, ∠4 + ∠1 + ∠5 = 180° ... ... (1)

But XPY ∥ QR and PQ, PR are transversals.

So, ∠4 = ∠2 and ∠5 = ∠3 (Pairs of alternate angles)

Substituting ∠4 and ∠5 in (1), we get

∠2 + ∠1 + ∠3 = 180°
That is,
∠1 + ∠2 + ∠3 = 180°

Recall that we have studied about the formation of an exterior angle of a triangle in the earlier classes (see Figure 6.36).

Figure 6.36

 

Side QR is produced to point S, ∠PRS is called an exterior ∠of triangle PQR.

Is ∠3 + ∠4 = 180°? (Why?) (1) Also, see that ∠1 + ∠2 + ∠3 = 180° (Why?) (2) From (1) and (2), we can see that

∠4 = ∠1 + ∠2.

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

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 opposite angles.

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

Example 7

In Figure 6.37, if QT is perpendicular to PR, ∠TQR = 40° and ∠SPR = 30°, find x and y.

Fig 6.37

 

Solution:

In triangle TQR, 90° + 40° + x = 180° (∠sum property of a triangle)

Therefore, x = 50°

Now, y = ∠SPR + x (Theorem 6.8)

Therefore, y = 30° + 50° = 80°

Example 8

In Figure 6.38, the sides AB and AC of Triangle ABC, are produced to points E and D respectively. If bisectors B O and C O of ∠CBE and ∠BCD respectively meet at point O, then prove that ∠BOC = 90° − ½ ∠BAC.

Figure 6.38

 

Solution :

Ray BO is the bisector of ∠CBE.

Therefore, ∠CBO =1/2 ∠CBE
=1/2(180° − y)
= 90° − y/ 2 ... ... (1)

Similarly, ray CO is the bisector of ∠BCD.

Therefore, ∠BCO =1/2 ∠BCD
=1/2(180° − z)
= 90° − z/2 ... ... (2)

In triangle B O C, ∠B O C + ∠BCO + ∠CBO = 180° ... ... (3)

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

∠BOC + 90° − z/2+ 90° − y/2 = 180°

So, ∠BOC =z/2 +y /2

or, ∠BOC = ½ (y + z) ... ... (4)

But x + y + z = 180° (∠sum property of a triangle)
Therefore, y + z = 180° − x

Therefore, (4) becomes

∠BOC = 1/2 (180° − x)
= 90° − x/2
= 90° − 1/2 ∠BAC

EXERCISE 6.3

Q1. In Figure 6.39, sides QP and RQ of triangle PQR are produced to points S and T respectively.

Figure 6.39

 

If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

A1. 65°

Q2. In Figure 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of triangle XYZ, find ∠OZY and ∠YOZ.

Fig 6.40

 

 

A2. 32°, 121°

Q3. In Figure 6.41, if AB ∥DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Figure 6.41

 

A3. 92°

Q4. In Figure 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Figure 6.42

 

A4. 60°

Q5. In Figure 6.43, if PQ is perpendicular to PS, PQ ∥ SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Fig 6.43

 

A5. 37°, 53°

Q6. In Figure 6.44, the side QR of triangle PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = ½ ∠QPR.

Fig 6.44

 

A6. Sum of the angles of the triangle PQR = Sum of the angles of triangle QTR and ∠PRS = ∠QPR + ∠PQR.