Finding $\frac{BD}{AE}$ when given that $AB = AC$ and $\angle BAC = 120^\circ$

Question

In triangle $ABC,$ $AB = AC$ and $\angle BAC = 120^\circ.$ If $D$ is the midpoint of $BC$ and $E$ is on $AB$ such that $DE \perp AB,$ find $\frac{BD}{AE}.$

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I was thinking that $\angle BAC$ could be used with LOC or such methods here, but I am unsure how to proceed from there. Can someone give me a hint?

Answer 1

Hint1: AD is also the height of the triangle or perpendicular to BC
Hint2: Since DE is perpendicular to AB, that means DE is a height
Hint3: ABD is a 30-60-90 triangle(AD is perpendicular to BC and ∠ABD = 1/2 * ∠BAD = 120/2 = 60)