Q. 193.7( 3 Votes )

# In Fig. 10.141, ABC is a triangle in which ∠B =2∠C. 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°

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)

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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