In figure, ABC is a triangle in which = 2 . D is a point on side BC such that AD bisects and AB = CD. BE is the bisector of . The measure of is:

In figure, ABC is a triangle in which = 2 . D is a point on side …

MCQ

In figure, $ABC$ is a triangle in which $\angle\text{B} = 2\angle\text{C}.$ $D$ is a point on side $BC$ such that $AD$ bisects $\angle\text{BAC}$ and $AB = CD. BE$ is the bisector of $\angle\text{B}.$ The measure of $\angle\text{BAC}$ is:

A
$95^\circ$
B
$73^\circ$
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$72^\circ$
D
$74^\circ$

✓

Answer

Correct option: C.

$72^\circ$

Given that $\triangle\text{ABC}$
BE is bisector of $\angle\text{B}$ and $AD$ is bisector of $\angle\text{BAC}$
$\angle\text{B} = 2\angle\text{C}$
By exterior angle theorem in triangle $ADC$
$\angle\text{ADB} = \angle\text{DAC} + \angle\text{C} ...\ \text{(i)}$
In $\triangle\text{ADB},$
$\angle\text{ABD} + \angle\text{BAD} + \angle\text{ADB} = 180^\circ$
$2\angle\text{C} + \angle\text{BAD} + \angle\text{DAC} + \angle\text{C} = 180^\circ$ [From $(i)]$
$3\angle\text{C} + \angle\text{BAC} = 180^\circ$
$\angle\text{BAC} = 180^\circ - 3\angle\text{C} ...\ \text{(ii)}$
Therefore,
$\text{AB = CD}$
$\angle\text{C} = \angle\text{DAC}$
$\angle\text{C} = \frac{1}{2}\angle\text{BAC}\ ...\ \text{(iii)}$
Putting value of Angle $C$ in $(ii)$, we get
$\angle\text{BAC} = 180^\circ - \frac{1}{2} \angle\text{BAC}$
$\angle\text{BAC} +\frac{3}{2} \angle\text{BAC} =180^\circ$
$\frac{5}{2} \angle\text{BAC} =180^\circ$
$\angle\text{BAC} = \frac{180\times2}{2}$
$=72^\circ$
$\angle\text{BAC} = 72^\circ$