In Fig.7.3, PQ > PR and QS and RS are the bisectors of ∠Q and ∠R, respectively. Show that SQ > SR.

Solution:

Given, PQR is a triangle.

PQ > PR

QS and RS are the bisectors of ∠Q and ∠R

We have to show that SQ > SR

Since QS is the bisector of ∠Q

∠PQS = ∠SQR

∠Q = ∠PQS + ∠SQR

∠Q = ∠SQR + ∠SQR

∠Q = 2∠SQR ----------------------------- (1)

Since RS is the bisector of ∠R

∠PRS = ∠SRQ

∠R = ∠PRS + ∠SRQ

∠R = ∠SRQ + ∠SRQ

∠R = 2∠SRQ ---------------------------- (2)

We know that in a triangle angle opposite the longer side is greater.

So, ∠R > ∠Q

From (1) and (2),

2∠SRQ > 2∠SQR

∠SRQ > ∠SQR ------------------------- (3)

We know that in a triangle side opposite to the greater angle is longer.

From (3),

SQ > SR

Therefore, it is proven that SQ is greater than SR.

✦ Try This: In the figure, AD∥BC and AD=BC. Then ΔABD≅ΔCDB by congruence postulate.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7

---

NCERT Exemplar Class 9 Maths Exercise 7.3 Sample Problem 3

In Fig.7.3, PQ > PR and QS and RS are the bisectors of ∠Q and ∠R, respectively. Show that SQ > SR

Summary:

The (interior) bisector of an angle, also called the internal angle bisector, is the line or line segment that divides the angle into two equal parts. In Fig.7.3, PQ > PR and QS and RS are the bisectors of ∠Q and ∠R, respectively. It is shown that SQ > SR

---

☛ Related Questions:

• ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE
• In Fig.7.4, D and E are points on side BC of a ∆ ABC such that BD = CE and AD = AE. Show that ∆ ABD . . . .
• CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). Show that ∆ ADE ≅ ∆ B . . . .