Find the measurement of line BD So I was trying to find the measurement of $BD$ I drew green lines to make myself some angles, the measurement $3$ is from the point A to C, If only I can line $AE$ or $CE$ then I will just use the cosine law to get $BE$ and $DE$, however I can't figure it out. 


Hint: Let's call $H_B$ the foot of the line from $B$ to $AC$ and $H_D$ the foot of the line from $D$ to $AC$
Then you have $AH_B=AB \cos 30$, $BH_B=AB \sin 30$  
and $CH_D=CD \cos 60$, $DH_D=CD \sin 60$
Thales gives us: $\dfrac{EB}{ED}=\dfrac{EH_B}{EH_D}=\dfrac{BH_B}{DH_D}$
But $H_BH_D=AC-AH_B-CH_D=H_BE+EH_D$
You can play with everything from now on...
EDIT: completing the work...
$H_BH_D=AC-AH_B-CH_D=3-AB \cos 30 -CD \cos 60=3-\dfrac{3\sqrt 3}{4}-2*\dfrac 12=\dfrac{8-3\sqrt 3}{4}$
$H_BE=\dfrac{H_BH_D}{1+\dfrac{DH_D}{BH_B}}=\dfrac{\dfrac{8-3\sqrt 3}{4}}{1+\dfrac{CD\sin 60}{AB \sin 30}}=\dfrac{\dfrac{8-3\sqrt 3}{4}}{1+\dfrac{4\sqrt 3}{3}}=\dfrac{24-9\sqrt 3}{12+16\sqrt 3}=\dfrac 34 \times \dfrac{8-3\sqrt 3}{3+4\sqrt 3}$
And $EH_D=\dfrac{4\sqrt 3}{3}EH_B=\sqrt 3 \times \dfrac{8-3\sqrt 3}{3+4\sqrt 3} $
$EB^2=BH_B^2+EH_B^2=\dfrac{9}{16}\times (1+(\dfrac{8-3\sqrt 3}{3+4\sqrt 3})^2)$
$ED^2=3\times (1+(\dfrac{8-3\sqrt 3}{3+4\sqrt 3})^2)$
$BD=\dfrac 14 \times (3+4\sqrt 3)\times \sqrt{1+(\dfrac{8-3\sqrt 3}{3+4\sqrt 3})^2}=\dfrac 12 \times \sqrt{37-6\sqrt 3}$
