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)
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