$\theta=\ ?$

Note that $\triangle ABC$ is isosceles that is $AB = AC$ . We have $\angle EBA = 180^\circ - 38^\circ - 33^\circ = 109^\circ$ , $\angle EAC = 180^\circ - 2 \theta - 33^\circ$ and $\angle ECA = 180^\circ - (180^\circ - 2 \theta - 33^\circ) - 19^\circ = 2\theta + 14^\circ$ . By sine rule ,

$\begin{cases} \dfrac {EA}{AB} = \dfrac {\sin 109^\circ}{\sin 38^\circ} & ...(1) \\ \dfrac {EA}{AC} = \dfrac {\sin (2\theta + 14^\circ)}{\sin 19^\circ} & ...(2) \end{cases}$

Since $AB=AC$ ,

$\begin{aligned} \frac {\sin (2\theta + 14^\circ)}{\sin 19^\circ} & = \frac {\blue{\sin 109^\circ}}{\red{\sin 38^\circ}} & \small \blue{\text{Note that }\sin (180^\circ - \phi)= \sin \phi} \\ \sin (2\theta + 14^\circ )& = \frac {\blue{\sin 71^\circ}}{\red{2\cos 19^\circ}} & \small \red{\text{and }\sin (2\phi) = 2\sin \phi \cos \phi} \\ \sin (2\theta + 14^\circ )& = \frac {\blue{\cos 19^\circ}}{2\cos 19^\circ} & \small \red{\text{also }\sin \phi = \cos (90^\circ - \phi)} \\ & = \frac 12 \\ \implies 2\theta + 14 & = 30^\circ \text{ or } 150^\circ \\ \theta & = 6^\circ \text{ or } \boxed{68}^\circ \end{aligned}$

Let $\angle {ABC}=\angle {ACB}=\theta$ . Then $|\overline {BC}|=2|\overline {AB}|\cos \theta$ . Let $\angle {AEB}=38\degree, \angle {BAE}=33\degree, \angle {AEC}=19\degree$ . Then $\dfrac{|\overline {AB}|}{|\overline {AE}|}=\dfrac{\sin 38\degree}{\sin 109\degree}=2\sin 19\degree, \dfrac{|\overline {AC}|}{|\overline {AE}|}=\dfrac{\sin 19\degree}{\sin (14\degree+2\theta)}$ . $|\overline {AB}|=|\overline {AC}|\implies \sin (14\degree+2\theta)=\dfrac{1}{2}$ . So $\theta$ can be $8\degree$ or $68\degree$ .