D is a point on the side BC of a ∆ABC such that AD bisects ∠BAC. Then
a. BD = CD
b. BA > BD
c. BD > BA
d. CD > CA

Solution:

It is given that

In ∆ ABC, AD is the angular bisector

AD meets BC at the point D

We know that

AD is the bisector of ∠BAC

∠BAD = ∠CAD

In ∆ ACD, external angle is ∠ADC

As the external angle of a triangle is greater than each internal opposite angle of the same triangle

∠ADB = ∠DAC + ∠DCA

Here ∠DAC = ∠BAD

∠ADB = ∠BAD + ∠DCA

i.e. ∠ADC > ∠BAD

BA > BD

Therefore, if D is a point on the side BC of a ∆ ABC such that AD bisects ∠BAC then BA > BD.

✦ Try This: S is a point on the side QR of a ∆ PQR such that PS bisects ∠QPR. Then a. QS = RS, b. QP > QS, c. QS > QP, d. RS > RP

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

NCERT Exemplar Class 9 Maths Exercise 7.1 Problem 6

D is a point on the side BC of a ∆ ABC such that AD bisects ∠BAC. Then, a. BD = CD , b. BA > BD , c. BD > BA , d. CD > CA

Summary:

D is a point on the side BC of a ∆ ABC such that AD bisects ∠BAC. Then BA > BD

☛ Related Questions:

• It is given that ∆ ABC ≅ ∆ FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then which of the following is . . . .
• Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle . . . .
• In ∆ PQR, if ∠R > ∠Q, then ,a. QR > PR, b. PQ > PR, c. PQ < PR, d. QR < PR