Trig Bash Solution. Let $\angle EAC = \alpha$. By the Angle Bisector Theorem and the Law of Sines we obtain that $$2\sin 70^\circ = \frac{AB}{BC}=\frac{AD}{DC} = \frac{AD}{DE} \cdot \frac{ED}{DC} = \frac{\sin (30^\circ - \alpha)}{\sin \alpha} \cdot \frac{\sin 70^\circ}{\sin 80^\circ}$$$$\Leftrightarrow$$$$2\sin \alpha \cos 10^\circ = \sin (30^\circ - \alpha)$$Note that $\alpha$ is uniquely determined and since $\alpha =10^\circ$ is work, we get the answer $\boxed{10^\circ}$

Clearly, $\angle BCA = 70^\circ$. Let $F$ be the reflection of $B$ in the line $AC$. We then have $\angle BAF = 2\angle BAC = 60^\circ$ and $AB=AF$, so $\Delta ABF$ is equilateral, thus $BF=AB$ We have $\angle BCF = 2\angle BCA = 140^\circ$ and since $BC=CF$, we get $\angle CBF = \angle CFB = 20^\circ$. Since $DE \parallel AB$, $\dfrac{CD}{DA}=\dfrac{CE}{EB}$. But, by the angle bisector theorem and since $AB=BF$ and $CF=BC$, we get $$\dfrac{CD}{DA}=\dfrac{BC}{AB}=\dfrac{CF}{BF},$$so $\dfrac{CD}{DA}=\dfrac{CF}{BF}$. By the converse of the angle bisector theorem it follows that $FE$ is the angle bisector of $\angle CFB$, so $\angle CFE = \angle BFE = 10^\circ$. Now we have $\angle EFA = \angle EFB + \angle BFA = 10^\circ + 60^\circ = 70^\circ = \angle ECA$. It follows that $AECF$ is cyclic, so $\angle EAC = \angle CFE = 10^\circ$.

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