Just The Other Way Of Asking It

Let us assume that $\triangle CDE$ is equilateral, then we check if $\angle AED = 15^\circ$ .

If $\triangle CDE$ is equilateral, then $\angle CDE = 60^\circ$ , $DA = DE$ and $\angle EAD = \angle AED$ . Let $\angle AED = \theta$ . Then, $180^\circ - \angle ADE = 2 \theta$ $\implies 180^\circ - 90^\circ - 60^\circ = 2 \theta$ $\implies 2 \theta = 30^\circ$ $\implies \theta = \angle AED = 15^\circ$ . Therefore, if $\triangle CDE$ is equilateral, $\angle AED = 15^\circ$ and the reverse is also true.

Proving this is quite tough. This is the only way I found to do so:

Since we know that $CE=DE$ , we can draw a straight line $EFG$ which is parallel to $AD$ , like this:

Notice that this line is also the symmetry line, line $EFG$ divides the figure in two.

Now, let $AD=a\implies AG =\dfrac{a}{2},\;DE=b,\;AE=c$ . From the properties of parallel lines , we know that $\angle DAE=\angle AEG=\theta$

From right-angle $\triangle AEG$ , we know that

$\sin \angle AEG = \dfrac{AG}{AE}\\ \sin \theta = \dfrac{\frac{a}{2}}{c}\\ \color{#3D99F6}{a=2c\sin \theta}$

From $\triangle ADE$ , we can use the sine rule :

$\dfrac{DA}{\sin \angle AED}=\dfrac{DE}{\sin \angle DAE}\\ \dfrac{a}{\sin15^{\circ}}=\dfrac{b}{\sin \theta}\\ b=\dfrac{\color{#3D99F6}{a}\sin\theta}{\sin15^{\circ}}=\dfrac{\color{#3D99F6}{2c\sin\theta}\sin\theta}{\sin15^{\circ}}\\ \color{#D61F06}{b=\dfrac{2c\sin^2\theta}{\sin15^{\circ}}}$

$\dfrac{DE}{\sin \angle DAE}=\dfrac{AE}{\sin \angle ADE}\\ \dfrac{\color{#D61F06}{b}}{\sin \theta}=\dfrac{c}{\sin (180^{\circ}-15^{\circ}-\theta)}\\ \color{#D61F06}{\dfrac{2\color{#20A900}{\cancel{\color{#D61F06}{c}}}\color{#20A900}{\cancel{\color{#D61F06}{\sin^2\theta}}\sin\theta}}{\sin15^{\circ}\color{#20A900}{\cancel{\color{#333333}{\sin \theta}}}}}=\dfrac{\color{#20A900}{\cancel{\color{#333333}{c}}}}{\color{#EC7300}{\sin (180^{\circ}-(15^{\circ}+\theta))}}\\ \sin15^{\circ}=\color{magenta}{2\sin\theta\sin(15^{\circ}+\theta)}\\ \sin15^{\circ}=\color{magenta}{\cos15^{\circ}-\cos(15^{\circ}+2\theta)}\\ \cos(15^{\circ}+2\theta)=\color{#69047E}{\cos15^{\circ}}-\sin15^{\circ}\\ \cos(15^{\circ}+2\theta)=\color{#69047E}{\sin75^{\circ}}-\sin15^{\circ}\\ \cos(15^{\circ}+2\theta)=\color{teal}{\sin(45^{\circ}+30^{\circ})-\sin(45^{\circ}-30^{\circ})}\\ \cos(15^{\circ}+2\theta)=\color{teal}{2\cos45^{\circ}\sin30^{\circ}}\\ \cos(15^{\circ}+2\theta)=2\left(\dfrac{1}{\sqrt{2}}\right)\left(\dfrac{1}{2}\right)\\ \cos(15^{\circ}+2\theta)=\dfrac{1}{\sqrt{2}}$

Now, we know that

$0^{\circ}<\theta<90^{\circ}\\ 0^{\circ}<2\theta<180^{\circ}\\ 15^{\circ}<15^{\circ}+2\theta<195^{\circ}$

Therefore, we can say that

$15^{\circ}+2\theta=45^{\circ} \implies \theta=\dfrac{45^{\circ}-15^{\circ}}{2}=15^{\circ}$

$\angle DAE=\angle AED$ , therefore, $\triangle ADE$ is an isosceles triangle where $AD=DE$

And since $AD=DC$ , $DE=EC$ , we know that

$AD=\color{olive}{DC=DE=EC}$

From this, we conclude that $\triangle CDE$ is equilateral. The answer is $\boxed{\text{Yes}}$

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Theorems used:

1. Reference angles: $0^{\circ}\leq\theta\leq180^{\circ},\;\color{#EC7300}{\sin(180^{\circ}-\theta)=\sin\theta}$
2. Product to sum trigonometric formula : $\color{magenta}{2\sin A\sin B = \cos(A-B)-\cos(A+B)}$
3. Complementary angles : $0^{\circ}\leq\theta\leq90^{\circ},\;\color{#69047E}{\cos\theta=\sin(90^{\circ}-\theta)}$
4. Sum to product trigonometric formula : $\color{teal}{\sin(A+B)-\sin(A-B)=2\cos A\sin B}$

Draw FE//AD and AG // DE

Let AF = 1 then GE = 2

∠EAG = 15 °

tan Θ = x , tan 15° = 2 - √3

tan (Θ + 15°) = x + 2 = $\frac{x + (2 - √3)}{1- x (2 - √3)}$

x + (2 - √3) = x(1- x (2 - √3)) + 2(1- x (2 - √3))

2 x² - √3 x² + 4x - 2√3 x - √3 = 0

(2 - √3) x² + (2 - √3) 2x - √3 = 0

x² + 2x - 3 - 2√3 = 0

x = √3

tan Θ =√3

Θ = 60°

then the triangle CDE is equilateral

Locate F on the E side of line AB such that triangle ABF is equilateral. The locus of all points G that are the vertex of a 15 degree angle AGD is an arc of the circle subtending chord AD in which angle AGD = 15 degrees can be inscribed. Since triangle ADF is isosceles and angle ADF = 150, it follows that F lies on this locus. Since F is equidistant from points C and D, it lies on the perpendicular bisector of segment CD. But point E also lies on the fore-mentioned locus and E is equidistant from points C and D, so F also lies on the perpendicular bisector of segment CD. Since the intersection of the locus arc and the perpendicular bisector of CD is unique on the E side of CD, it follows that E = F. Thus triangle CDE is equilateral.