Guessing the nature of a triangle if one angle is 60 degrees

Question

In triangle ABC, AB=12cm, angle B=60 degrees, the perpendicular from A to BC meets it at D. The bisector of angle ABC meets AD at E. Then E divides AD in the ratio $3:1,6:1,1:1,2:1$?

I assumed it be an equlateral triangle. So, median, altitude and angle bisector are all same. So, answer is $2:1$.

My question is what are the chances that this triangle is not equilateral? If this triangle is always equilateral then how can we modify it to make it scalene, keeping the angle $B$ $60$ degrees always?

Answer 1

Just use trigonometry to find $AD = 6 \sqrt{3}, ED = \frac{6}{\sqrt{3}}$ and hence $\frac{AE}{ED} = 2:1$. There is no need to assume that the triangle is equilateral.

Answer 2

Logic: To get the ratio in which $E$ divides $AD$, you need to calculate any $2$ of the lengths $AE,ED$ or $AD$. $$AD = 12\sin \frac{\pi}{3} = 6\sqrt{3}$$ $$BD = 12\cos \frac{\pi}{3} = 6$$ $$ED = BD\tan \frac{\pi}{6} = \frac{6}{\sqrt{3}}$$ So you can conclude that, $AD = 3ED \implies AE+ED=3ED \implies AE=2ED$.Thus: $$\frac{AE}{ED} = \frac{2}{1}$$

Answer 3

In a triangle, the bisector of an angle divides the opposite side in the same ratio as the ratio of the leg lengths of the bisected angle. The cosine of angle ABC, which we know to be $60°$, is $\frac12$, so $\overline{BD}$ is $6cm$. $\overline{AE}$ is twice the size of $\overline {DE}$.
For a triangle having an angle of 60°, the chances are simply EXCELLENT that it is nevertheless scalene!