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The semiperimeter of $\triangle DXY$ is equal to $1$, which implies that the $D$-excircle of $\triangle DXY$ is tangent to $CD$ at $C$ and $DE$ at $E$ since $CD = DE = 1$. But since $\angle DCA = \angle DEA = 90^{\circ}$ (they both form $30-60-90$ triangles), this must be the circle centered at $A$ passing through $C$ and $E$. So, if we let $I$ be the incenter of $\triangle DXY$, then $XIYA$ must be cyclic by the Incenter-Excenter lemma, and hence $\angle XAY = 180^{\circ} - \angle XIY = 180^{\circ} - (90^{\circ} + \tfrac{1}{2} \angle XDY) = 30^{\circ}$.

Alternatively, we can cheese by assuming $DX=DY$, then we have that $x+x+x\sqrt{3}=2$ where $x=DX=DY$. This gives $XY=4-2\sqrt{3}$. After this it is easy to find distances and use LoC to compute the value of $\angle{XAY}$.

S.Das93 wrote: Alternatively, we can cheese by assuming $DX=DY$ Another way to cheese is to assume $D \equiv X$ and $E \equiv Y$, which completely trivializes the problem, though the nice answer also hints at $A$ being a special point in relation to $\triangle DXY$

TRIGONOMETRY FTW!! It's easy to see that $AC=AE$ by congruence. Now by law of cossines on $\triangle AFE$ with have $AE^2=1^2+1^2-2*1*1cos120^{\circ} \implies AC=AE=\sqrt{3}.$ Let $DY=b, DX=a$ with this we have $XY=2-(a+b), EY=1-b, XC=1-a$ , and we can see that the triangles $AEY$ and $ACX$ are right so by the well known pythagoras: $AY=\sqrt{4+b^2-2b}$ and $AX=\sqrt{4+a^2-2a}$. Now by law of cossines on $\triangle XAY$: $[(2-(a+b)]^2=(\sqrt{4+b^2-2b})^2 + (\sqrt{4+a^2-2a})^2 - 2\sqrt{(4+b^2-2b)(4+a^2-2a)} cos\angle XAY$. $\implies \cancel{4}+\cancel {a^2+b^2}+2ab-\cancel{4(a+b)}=\cancel{a^2+b^2-2(a+b)+8}-2\sqrt{(4+b^2-2b)(4+a^2-2a)} cos\angle XAY$ $\implies 2ab-2a-2b-4= - 2\sqrt{(4+b^2-2b)(4+a^2-2a)} cos\angle XAY$. ( dividing both sides by $-2$ and isolating $cos \angle XAY$): $\implies cos \angle XAY=\frac{a+b+2-ab}{\sqrt{(4+b^2-2b)(4+a^2-2a)}}$ Now lets find $sen \angle XAY$ , and to do that, we can calculate the area of the $\triangle AXY$ in two different ways: $[AXY]=\frac{AY*AX*sen \angle XAY}{2}=[ABCDEF]-[2(\frac{AF*FE*cos 120^{\circ}}{2})+\frac{EY*AE}{2}+\frac{AC*XC}{2}+\frac{DX*DY*cos120^{\circ}}{2}]$ $\implies \frac{\sqrt{(4+b^2-2b)(4+a^2-2a)} sen\angle XAY}{2}=\frac{6\sqrt{3}}{4}-(\frac{ab\sqrt{3}+2\sqrt{3}+2\sqrt{3}(2-a-b)}{4}$ $\implies \sqrt{(4+b^2-2b)(4+a^2-2a)} sen \angle XAY=\frac{\sqrt{3}(4-ab-2(2-a-b))}{2}$ $\implies sen \angle XAY=\frac{\sqrt{3}(4-ab-2(2-a-b))}{2(\sqrt{(4+b^2-2b)(4+a^2-2a)})}$ Now notice that $\frac {sen \angle XAY}{cos \angle XAY}=tan \angle XAY=\frac{\sqrt{3}(4-ab-2(2-a-b))}{2(a+b+2-b)}$ $\implies tan \angle XAY=\frac{2(a+b)-ab)\sqrt{3}}{2a+2b-2ab+4}$ Using Law of cossines on the $\triangle DXY$ we get that $(2-a-b)^2=a^2+b^2-ab*cos120^{\circ}$ $\implies 4+\cancel{a^2+b^2}+2ab-4a--4b=\cancel{a^2+b^2}+ab$ $\implies 4-4a-4b=-ab$ So now $\sqrt{3}(2a+2b-ab)=\sqrt{3}(2a+2b+4-4a-4b=\sqrt{3}(-2a-2b+4)$ and $2a+2b-2ab+4=2a+2b+8-8a-8b+4=3(-2a-2b+4) $ which means that $tan \angle XAY=\frac{\cancel{(-2a-2b+4)}\sqrt{3}}{\cancel{(-2a-2b+4)}3} \implies tan \angle XAY=\frac{\sqrt{3}}{3}$. We know that $\angle XAY$ is acute so $\angle XAY=30^{\circ}$