Not Rotating Equilateral Triangles

$\angle ACE=5$ calculated for quadrilateral $ACEB$ .

Triangles $BCE$ and $ACD$ have two sides of the same length: $BC=AC, EC=CD$ . Their corresponding angles are different though: $\angle BCE=45, \angle ACD=60-5=55$ . We can easily create a triangle congruent to $\triangle ACD$ by the following construction.

Let $F$ denote point of intersection of $BE$ and $AC$ . $\angle BFC=90$ . Let $G$ denote reflection of $E$ about $AC$ . $\angle FCE=\angle FCG=5$ . Triangles $BGC$ and $ACD$ are congruent because $\angle BCG=\angle ACD=55$ , therefore $\angle ADC=\angle BGC =85$ . $\angle ADE=\angle ADC-60=\boxed{25}$ .

We note that $\angle EBC = 180^\circ - \angle BEC - \angle ECB = 180^\circ - 95^\circ - 45^\circ = 40^\circ$ . Therefore, $\angle ABC =$ $\angle ABE + \angle EBC =$ $25^\circ + 40^\circ =$ $65^\circ$ . This means that $\angle BAC =$ $\angle ABC = 65^\circ$ and that $AC = BC$ .

Now, let $CD=DE=EC=1$ and $AC=BC = a$ . Considering sine rule in $\triangle ECB$ :

$\begin{aligned} \frac {BC}{\sin \angle BEC} & = \frac {EC}{\sin \angle ECB} \\ \frac a{\sin 95^\circ} & = \frac 1{\sin 40^\circ} \\ \implies a & = \frac {\sin 95^\circ}{\sin 40^\circ} & ... (1) \end{aligned}$

Let $\angle ADE = \theta$ . Then $\angle DAC = 180^\circ - \angle ADC - \angle DCA =$ $180^\circ - (\theta + 60^\circ) - 55^\circ =$ $65^\circ - \theta$ . Considering sine rule in $\triangle ADC$ :

$\begin{aligned} \frac {AC}{\sin \angle ADC} & = \frac {CD}{\sin \angle DAC} \\ \frac a{\sin (\theta + 60^\circ)} & = \frac 1{\sin (65^\circ-\theta)} \\ \implies a & = \frac {\sin (\theta + 60^\circ)}{\sin (65^\circ-\theta)} & ... (2) \end{aligned}$

Then we have:

$\begin{aligned} (2)=(1): \quad \frac {\sin (\theta + 60^\circ)}{\sin (65^\circ-\theta)} & = \frac {\color{#3D99F6}\sin 95^\circ}{\sin 40^\circ} & \small \color{#3D99F6} \text{Note that } \sin (180^\circ - x) = \sin x \\ & = \frac {\color{#3D99F6}\sin 85^\circ}{\sin 40^\circ} \\ & = \frac {\sin (25^\circ + 60^\circ)}{\sin (65^\circ - 25^\circ)} \\ \implies \theta & = \boxed{25}^\circ \end{aligned}$

$Note~that~\angle s~~EBC=40^o,~~~CDA=60+\alpha,~~~DAC=65-\alpha,\\ ABC=65^o,~~~but~also~BAC=65^o.~~\implies~AC=BC.\\ WLOG~CD=DC=1.~~~Using~~Sin~~Law:- \\ \Delta~EBC,~~~AC=BC=\dfrac{Sin95}{Sin40}*1..........(1)\\ \Delta~DAC,~~~AC=\dfrac{Sin(60+\alpha)}{Sin(65-\alpha)}*1..........(2)\\ Equating ~(1)~~and~~(2)~~,~expanding~Sin(A+B),~we~get,\\ (Sin95*Sin65 - Sin40*Sin60)*Cos\alpha=(Sin40*Cos60 + Sin95*Cos65)*Sin\alpha. \\ \therefore~~Replacing~values ~of~Sins~,~Coses~and~ ArcTan,~ and~ simplifying~~~ \alpha~=~\Large~~\color{#D61F06}{25}.$

I am inviting proofs for this one.