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°
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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