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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 $\frac{BD}{AE}.$

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?

questionasker
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    $BDA$, $BDE$, $DAE$ are all 30-60-90 triangles. – Intelligenti pauca Sep 28 '20 at 19:44
  • You can also put all the points in a coordinate system. When you put A at (0, 0), B at (1, 0) then C, D and E can be easily computed. (In that order) You should of course add an explanation why this scaling and rotating the triangle is allowed to your solution. – Rolf Kreibaum Sep 28 '20 at 19:53
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    I've got it, thanks! @Intelligentipauca's comment was very helpful. – questionasker Sep 28 '20 at 19:54

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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)

lier wu
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