Question 1106412
<pre><font size = 4><b>
{{{drawing(400,1000/3,-3,3,-1,4,
locate(-.02,3.7,A),locate(-2,0,B),locate(2,0,C), locate(-1,0,E),locate(0,0,D),
triangle(-2,0,2,0,0,2sqrt(3)),
green(line(0,0,0,2sqrt(3))), 

locate(-.5,0,x), locate(-1.5,0,x),locate(1,0,2x),

locate(.06,1.6,2x*sqrt(3)),
locate(1.1, 1.8,4x), locate(-1.3, 1.8,4x), 
red(line(-1,0,0,2sqrt(3)))  )}}}
<pre>
Let BE = ED = x.  Then
BD = 2x = DC.  Then
BC = 4x = AB

By using the Pythagorean theorem on right triangle ADC,
AD = {{{2x*sqrt(3)}}}

EC = ED + DC = x + 2x = 3x

Area of triangle AEC = {{{expr(1/2)*base*altitude}}}{{{""=""}}}{{{expr(1/2)EC*AD}}}{{{""=""}}}{{{expr(1/2)(3x)(2x*sqrt(3))}}}{{{""=""}}}{{{3x^2*sqrt(3)}}}

Since the area of triangle AEC = {{{27*sqrt(3)}}}, we have
the equation:

{{{3x^2*sqrt(3)}}}{{{""=""}}}{{{27*sqrt(3)}}}

Divide both sides by {{{3sqrt(3)}}}

{{{x^2}}}{{{""=""}}}{{{9}}}

{{{x}}}{{{""=""}}}{{{3}}}

So AB = 4x = 4(3) = 12

Edwin</pre></font></b>