b？ = AB？ + BC？ -2AB * BC * cos∠ABC = 12+64-2 * 2√3 * 8 *( 1/2)= 76- 16√3

b = 2√( 19 - 4√3)

Then use sine theorem: BC/sin∠BAC = AC/sin∠ABC.

sin∠BAC = BC * sin∠ABC/AC = 8 *(√3/2)/[2 √( 19-4√3)]= 2√3/√( 19-4√3)

Reuse cosine theorem: cos∠BAC = (AC? + AB？ -BC? )/(2AC * AB)=(76- 16√3+ 12-64)/[2 * 2 √( 19-4√3)* 2√3]=(√3-2)/√( 19-4√3)

In the triangle ABD BD? = AB？ + AD？ - 2AB*AD*cos∠BAD ( 1)

Here AB is known, and so is AD=AC. ∠BAD = 360-∠BAC-∠DAC = 360-(∠BAC+ 120)

cos∠BAD = cos[360-(∠BAC+ 120)]= cos[-(∠BAC+ 120)]= cos(∠BAC+ 120)

= cos∠BAC * cos 120-sin∠BAC * sin 120

Because the sine and cosine of ∠BAC are known, we can find out cos∠BAD, and then use (1) to find out BD. The process is a bit cumbersome, but it can certainly be done.