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]

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 = 180^o

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

3 ∠C + ∠BAC = 180^o

∠BAC = 180^o – 3 ∠C (ii)

Therefore,

AB = CD

∠C = ∠DAC

∠C = 1/2 ∠BAC (iii)

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

∠BAC = 180^o – 1/2 ∠BAC

∠BAC + ∠BAC = 180^o

∠BAC = 180^o

∠BAC =

= 72^o

∠BAC = 72^o