1. Let $C'$ be a point on the ray $EB$ such that $EC' = EC$. From the congruence of triangles $AEC'$ and $DEC$, we get $AC' = DC = AB$, but $C' \equiv B$, so it follows that $180^\circ = \angle ABD + \angle AC'D = \angle ABD + \angle ACD$. This means that the rays $AB$ and $DC$ intersect at some point $F$ (since $\angle ABC + \angle BCD > 180^\circ$) and that the quadrilateral $BEFC$ is cyclic. Now,

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$\angle EFA = \angle ECB = \angle EAF$, so $EF = EA = ED$, i.e., $E$ is the center of the circumcircle of $\triangle ADF$. Finally, $2 \angle AFD = \angle AED = \angle BEC = 180^\circ - \angle AFD$, so $\angle AFD = 60^\circ$ and $\angle BAD + \angle ADC = 120^\circ$.

Second Solution. Let $AB = BC = CD = 1$, $\angle ACB = x$, and $\angle DBC = y$. We have $AC = 2 \cos x$ and $CE = \frac{\sin y}{\sin (x+y)}$ by the Law of Sines in $\triangle BCE$, so $AE = 2 \cos x - \frac{\sin y}{\sin (x+y)} = \frac{2 \sin (x+y) \cos x - \sin y}{\sin (x+y)} = \frac{\sin (2x + y)}{\sin (x+y)}$. Similarly, $DE = \frac{\sin (x + 2y)}{\sin (x+y)}$, so from the condition $AE = DE$ it follows that $0 = \sin (2x + y) - \sin (x + 2y) = 2 \sin \frac{x - y}{2} \cos \frac{3x + 3y}{2}$. Since $x \neq y$ and $x, y < 90^\circ$, it must be that $\frac{3x + 3y}{2} = 90^\circ$, i.e., $x + y = 60^\circ$. Finally, $\angle ABC + \angle BCD = 360^\circ - 2x - 2y = 240^\circ$, from which the statement immediately follows.