Question : In a triangle ${ABC}, {D}$ is a point on ${BC}$ such that $\frac{A B}{A C}=\frac{B D}{D C}$. If $\angle B=68^{\circ}$ and $\angle C=52^{\circ}$, then measure of $\angle B A D$ is equal to:
Option 1: $50^{\circ}$
Option 2: $40^{\circ}$
Option 3: $60^{\circ}$
Option 4: $30^{\circ}$
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Correct Answer: $30^{\circ}$
Solution :
Given: In a triangle ${ABC}, {D}$ is a point on ${BC}$ such that $\frac{A B}{A C}=\frac{B D}{D C}$.
The angle bisector theorem states that an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.
The $\angle B=68^{\circ}$ and $\angle C=52^{\circ}$.
The sum of all the angles in a triangle = $180^{\circ}$.
⇒ $\angle A + \angle B + \angle C = 180^{\circ}$
⇒ $\angle A +68^{\circ}+52^{\circ} = 180^{\circ}$
⇒ $\angle A +120^{\circ} = 180^{\circ}$
⇒ $\angle A =180^{\circ} – 120^{\circ}$
⇒ $\angle A =60^{\circ}$
The measure of $\angle B A D=\frac{60^{\circ}}{2}=30^{\circ}$.
Hence, the correct answer is $30^{\circ}$.
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