Q. 193.7( 3 Votes )

In Fig. 10.141, ABC is a triangle in which B =2C. D is a point on side such that AD bisects BAC and AB=CD. BE is the bisector of B. The measure of BAC is



[Hint: Δ ABE Δ DCE]
A. 72°

B. 73°

C. 74°

D. 95°

Answer :

Given that ABC

BE is bisector of Band AD is bisector of BAC


B = 2 C


By exterior angle theorem in triangle ADC


ADB = DAC + C (i)


In ADB,


ABD + BAD + ADB = 180o


2 C + BAD + DAC + C = 180o [From (i)]


3 C + BAC = 180o


BAC = 180o – 3 C (ii)


Therefore,


AB = CD


C = DAC


C = 1/2 BAC (iii)


Putting value of Angle C in (ii), we get


BAC = 180o – 1/2 BAC


BAC + BAC = 180o


BAC = 180o


BAC =


= 72o


BAC = 72o

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