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 = 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}

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