ABC is an isosceles triangle with AB = AC. The vertex angle A measures eight times the measure of a base angle. AD is the angle bisector of vertex angle A. What is the measure of the angle BAD?

Given : ABC is an isosceles triangle, with AB=AC, let $$\angle$$ ABC $$=\angle$$ ACB = $$x$$

=> $$\angle$$ BAC = $$8x$$

Also, AD is angle bisector, => $$\angle$$ BAD $$=\angle$$ CAD = $$4x$$

The angle bisector of an isosceles triangle is perpendicular bisector to the base.

=> $$\angle$$ BDA $$=\angle$$ CDA = $$90^\circ$$

Thus, in right triangle ABD, => $$\angle$$ ABD + $$\angle$$ ADB + $$\angle$$ BAD = $$180^\circ$$

=> $$x+4x=180-90=90^\circ$$

=> $$x=\frac{90}{5}=18^\circ$$

$$\therefore$$ $$\angle$$ BAD = $$4\times18=72^\circ$$

=> Ans - (C)