Ratio between weird lengths

The angles around A must add up to $360^\circ$ :

$60^\circ+60^\circ+90^\circ+EAF=360^\circ -> EAF=150^\circ$

$BAE=90^\circ+60^\circ=150^\circ$

This means that $EF=EB$ since they have two congruent sides and a congruent angle.

By a $60^\circ$ rotation around $C$ $CE$ will map to $CA$ and $CB$ will map to $CD$ .

Therefore $EB$ must map to $AD$ . But since we already showed that $EF=EB$ this means that $EF=AD$ .

Since the lengths are the same the ratio $\frac{AD}{EF}$ is $1$

Let $BC = 1$ and $\angle BCA = theta$ . Then $AC = \cos \theta$ and $AB = \sin \theta$ . Let the foot of the altitude from $A$ to the extension of $DC$ be $G$ . Applying the Pythagorean theorem on the right $\triangle ADG$ :

$\begin{aligned} AD^2 & = DG^2 + AG^2 \\ & = (DC+CG)^2 + AG^2 \\ & = (DC+AC\cos {\color{#3D99F6}\angle ACG})^2 + (AC \sin {\color{#3D99F6}\angle ACG})^2 \quad \quad \small \color{#3D99F6} \text{Note that }\angle ACG = 180^\circ - 60^\circ - \theta \\ & = (1 + \cos \theta \cos {\color{#3D99F6} (120^\circ - \theta)})^2 + (\cos \theta \sin {\color{#3D99F6} (120^\circ - \theta)})^2 \\ & = 1 + 2\cos \theta \cos (120^\circ - \theta) + \cos^2 \theta (\cos^2 (120^\circ - \theta) + \sin^2 (120^\circ - \theta)) \\ & = 1 + 2\cos \theta \left(-\frac 12 \cos \theta + \frac {\sqrt 3}2\sin \theta \right) + \cos^2 \theta \\ & = 1 + \sqrt 3 \sin \theta \cos \theta \\ \implies AD & = \sqrt{1 + \sqrt 3 \sin \theta \cos \theta} \end{aligned}$

Using cosine rule on $\triangle AEF$ , we have:

$\begin{aligned} EF^2 & = AE^2 + AF^2 - 2AE \cdot AF \cos \color{#3D99F6} \angle EAF & \small \color{#3D99F6} \text{Note that }\angle EAF = 360^\circ - 60^\circ - 90^\circ - 60^\circ \\ & = \cos^2 \theta + \sin^2 \theta - 2sin \theta \cos \theta \cos \color{#3D99F6} 150^\circ \\ & = 1 + \sqrt 3 \sin \theta \cos \theta \\ \implies EF & = \sqrt{1 + \sqrt 3 \sin \theta \cos \theta} \end{aligned}$

Therefore, $\dfrac {AD}{EF} = \dfrac {\sqrt{1 + \sqrt 3 \sin \theta \cos \theta}}{\sqrt{1 + \sqrt 3 \sin \theta \cos \theta}} = \boxed {1.000}$ .

I arbitarily set the AB distance to $1$ as we are only interested in scaled values.I rotated the figure so that point A was the origin and point $c$ was on the positive y axis. $dx$ and $dy$ are the $x$ and $y$ coordinates of point D. The second line is the requirement that the ABC triangle be a right triangle with with the right angle being point A. The third and fourth lines are the requirement that triangle BCD be equilateral. The other two triangles are equilateral by construction. $\text{Reduce}\left[c>0\land \text{dx}\geq 1\land \text{dy}\geq 0\land c^2-\text{dx}^2+1\geq 0\land \\ \text{SquaredEuclideanDistance}[\{0,0\},\{0,c\}]+\text{SquaredEuclideanDistance}[\{0,0\},\{1,0\}]=\text{SquaredEuclideanDistance}[\{0,c\},\{1,0\}]\land \\ \text{SquaredEuclideanDistance}[\{0,c\},\{1,0\}]=\text{SquaredEuclideanDistance}[\{0,c\},\{\text{dx},\text{dy}\}]\land \\ \text{SquaredEuclideanDistance}[\{0,c\},\{1,0\}]=\text{SquaredEuclideanDistance}[\{1,0\},\{\text{dx},\text{dy}\}],\\ \{b,c,\text{dx},\text{dy}\},\mathbb{R}\right]$ $\left(c=\frac{1}{\sqrt{3}}\land \text{dx}=1\land \text{dy}=\frac{2}{\sqrt{3}}\right)\lor \\ \left(\frac{1}{\sqrt{3}}<c<\sqrt{3}\land \text{dx}=\frac{1}{2} \sqrt{3} \sqrt{c^2}+\frac{1}{2}\land \text{dy}=\sqrt{c^2-\text{dx}^2+1}+c\right)\lor \\ \left(c\geq \sqrt{3}\land \text{dx}=\frac{1}{2} \sqrt{3} \sqrt{c^2}+\frac{1}{2}\land \text{dy}=c-\sqrt{c^2-\text{dx}^2+1}\right)$

Back-substituting the three solution ranges into the distance ratio formula gives $1$ in all three cases. $\frac{\text{EuclideanDistance}[\{0,0\},\{\text{dx},\text{dy}\}]}{\text{EuclideanDistance}\left[\left\{\frac{1}{2},-\frac{\sqrt{3}}{2}\right\},c \left\{-\frac{\sqrt{3}}{2},\frac{1}{2}\right\}\right]}$

Since the solution is valid for all value of AB, let $AB \to 0$ . Then it is trivial to see that you have two back to equilateral triangles, so FE = AD.

Given that

$\angle CAB = 90°, \angle ABC = x$

$sin (x) = \frac {AC}{BC}, cos (x) = \frac {AB}{BC}$

and

$\angle EAC = \angle FAB = 60°$

$\therefore \angle EAF = 150°$

In triangle AEF, by cosine rule

$EF^2 = AE^2 + AF^2 - 2*AE*AF*cos(150°)$

In triangle AEB, AE = AB

similarly in triangle ACF, AF = AC

$\Rightarrow EF^2 = AB^2 + AC^2 + \sqrt{3} *AB*AC$

$\Rightarrow EF^2 = BC^2 + \sqrt{3} *AB*AC ---(1)$

In triangle ADB, by cosine rule

$AD^2 = AB^2 + DB^2 - 2*AB*DB*cos (60° + x)$

$cos (60 + x) = cos (60) * cos (x) - sin (60) * sin (x)$

$cos (60 + x) = \frac {AB}{2 * BC} - \frac {\sqrt{3} * AC}{2 * BC} = \frac{AB - \sqrt{3} * AC}{2 * BC}$

In triangle DBC, BC = DB

$\Rightarrow AD^2 = AB^2 + BC^2 - 2*AB*BC* \frac{AB - \sqrt{3} * AC}{2 * BC}$

$\Rightarrow AD^2 = AB^2 + BC^2 - AB*(AB - \sqrt{3} * AC)$

$\Rightarrow AD^2 = BC^2 + \sqrt{3} * AB * AC ---(2)$

(1) = (2)

$\therefore \frac{AD}{EF} = 1$