it is not $30^\circ$

Some of the easily obtainable angles are listed in the image above.

Let’s set $EF=1$ . From repeated use of the law of sines in triangles $EFC, CFB$ , and $BFD$ we get

$DF=a=\frac{\sin65}{\sin45}\frac{\sin34}{\sin36}\frac{\sin43}{\sin67}=0.9034$

From law of cosines in $\triangle DEF$ the side $DE=b=\sqrt{1+a^2-2a\cos110^\circ}=1.56$

So the angle $\alpha$ , also from the law of cosines in the same triangle

$\alpha=\arccos(\frac{a^2+b^2-1}{2ab})\approx\boxed{37^\circ}$

Since $\triangle ABC$ is isosceles, $\angle ABC = \angle ACB = \dfrac {180^\circ - 22^\circ}2=79^\circ$ . Then $\angle BEC = 180^\circ - 36^\circ - 79^\circ = 65^\circ$ and $\angle BDC = 180^\circ - 34^\circ - 79^\circ = 67^\circ$ .

Let $BC=1$ and use sine rule on $\triangle BEC$ :

$\begin{aligned} \frac {BC}{\sin \angle BEC} & = \frac {EC}{\sin \angle EBC} \\ \frac 1{\sin 65^\circ} & = \frac {EC}{\sin 36^\circ} \\ \implies EC & = \frac {\sin 36^\circ}{\sin 65^\circ} \end{aligned}$

Using sine rule on $\triangle BDC$ :

$\begin{aligned} \frac {BC}{\sin \angle BDC} & = \frac {DC}{\sin \angle DBC} \\ \frac 1{\sin 67^\circ} & = \frac {DC}{\sin 79^\circ} \\ \implies DC & = \frac {\sin 79^\circ}{\sin 67^\circ} \end{aligned}$

Let $\angle CDE = \alpha$ . We note that $angle DCE = 79^\circ - 34^\circ = 45^\circ$ and $\angle DEC = 180^\circ - 45^\circ - \alpha = 135^\circ - \alpha$ . Using sine rule on $\triangle CDE$ :

$\begin{aligned} \frac {\sin \angle CDE}{EC} & = \frac {\sin \angle DEC}{DC} \\ \frac {\sin \alpha}{EC} & = \frac {\color{#3D99F6}\sin (135^\circ - \alpha)}{DC} & \small \color{#3D99F6} \text{Note that }\sin (180^\circ - x) = \sin x \\ DC \sin \alpha & = EC \color{#3D99F6} \sin (45^\circ + \alpha) \\ DC \sin \alpha & = \frac {EC}{\sqrt 2} (\sin \alpha + \cos \alpha) \\ \implies \tan \alpha & = \frac {EC}{\sqrt 2 DC - EC} \\ & = \frac 1{\frac {\sqrt 2 DC}{EC} - 1} \\ & = \frac 1{\frac {\sqrt 2 \sin 79^\circ \sin 65^\circ}{\sin 67^\circ \sin 36^\circ}-1} \\ & \approx 0.646378022 \\ \implies \alpha & \approx \boxed{37^\circ} \end{aligned}$