In △ABC, ∠A= 60$^∘ $, BC=12, BD⊥AC, CE⊥AB and∠DBC = 3∠ECB. Find EC in the form $a(\sqrt{b}+\sqrt{c})$ Given that $m \angle A= 60^\circ$, $BC=12$ units, $\overline{BD}\perp\overline{AC}$, $\overline{CE} \perp \overline{AB}$, and $m \angle DBC = 3m \angle ECB$, the length of segment $EC$ can be expressed in the form $a(\sqrt{b}+\sqrt{c})$ units where $b$ and $c$ have no perfect-square factors. What is the value of $a+b+c$?

Let the intersection of $EC$ and $BD$ be $F$. I have figured out that $\triangle EFB$ and $\triangle DFC$ are both $30-60-90$ triangles, and that $\triangle BDC$ is a $45-45-90$ triangle. Along with this, I have figured out all other angles. However, I can't figure out the next step. Can someone help? Thanks!

Let $\angle ECB = y$, so $\angle DBC=3y$. From $\triangle AEC$ we have $\angle ACE = 180^\circ-60^\circ-90^\circ= 30^\circ$.
Now let $EC$ and $BD$ intersect at $F$. $\angle BFE=\angle DFC$ by vertical angles and $\angle BEF=\angle CDF=90^\circ$, so $\angle FBE=\angle FCD$, which is equal to 30 degrees. Now summing the angles in $\triangle ABC$, we have $60^\circ+30^\circ+3y+y+30^\circ=180$, solving yields $4y=60$ so $y=15$ and we see $\triangle BDC$ is a 45-45-90 triangle. Also, $\triangle ABD$ is a 30-60-90 triangle.
Let $ AD = x$, so $AB = 2x$ and $DB = DC = x\sqrt{3}$. $BC = x\sqrt{3}\sqrt{2} = x\sqrt{6}$. We are given that this equals 12, so we find $x = 12/\sqrt{6} = 2\sqrt{6}$. It follows that the area of $\triangle ABC$ can be found via[(1/2)(AC)(BD)=(1/2)(x+x\sqrt{3})(x\sqrt{3})=12\sqrt{3}+36.]To find $EC$, notice that the area of $\triangle ABC$ can also be written as $(1/2)(AB)(EC)$. Thus,[(1/2)(4\sqrt{6})(EC)=12\sqrt{3}+36 \Rightarrow EC = 3(\sqrt{2}+\sqrt{6}).]Hence $a=3$, $b=2$, and $c=6$, so $a+b+c=\boxed{11}$.
