Geometry question involving the length of a chord Two chords $AB$ and $AC$ are drawn inside a circle with diameter $AD$. The angle $BAC = 60$, $AB = 24cm$, $EC = 3cm$, and $BE$ and $AC$ are perpendicular. What is the length of the chord $BD$?
Here's what I've tried:
$ABE = 30$ which implies $AE = 12cm$ and therefore $BE = 12\sqrt3cm$ and $BC = 21cm$. Call the intersection of $EB$ and $AD$ point $O$. So we have $AEO$ is similar to $ACD$  but I don't know where to go from here.
Thank you so much in advance!

Another approach:

$\angle ABE=30^o$
$\angle ABD=90^o$
because it is opposite to diameter AD.Therefore:
$\angle OBD=90-30=60^o$
$DF||EC$. SO :$DF=EC=3$
In right angled triangle BFD we have:
$BD=\frac{3}{Sin(60)}=\frac{3}{\sqrt 3/2}=2\sqrt 3$
