Solve for the angle x? The problem seems really impossible to me, despite solving many problems on triangles. So I hope that you can solve it.

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 A: 
Unfortunately, some case analyses and contradiction things must be done and I wasn't able to come up with any nice and clean way.
First create triangle $ACG$ congruent to $ABC$ and label $F$ as shown in the picture. Mirror $D$ along axis $AC$ to point $P$. Connect necessary points as shown in the picture.
In trapezoid $ADCG$, triangles $ACG$ and $DGC$ are symmetric so congruent as well. So $DGC$ is also a triangle congruent to $ABC$ where angle $\angle DGC=x$
At the same time, $PA = AD = EC, AG = CD, \angle PAG = 90 - {x\over 2} - x = \angle ECD \implies \triangle PAG \cong \triangle ECD$. 
Furthermore it's easy to see $DE=PG$ and $\triangle DEG \cong \triangle GPC$.
Now let $\angle GDE = \angle CGP = a$.
With some angle tracing (with the known congruences every single angle can be expressed in terms of x), $a = 40 - {x\over 2}$, and $\angle DEP = 280 - 3x$.
Case $1$: $EP$ is not parallel to $DG$ and intersect $DG$ on the extension ray from $D$ to $G$. Then this means $\angle GDE + \angle DEP < 180$ which means $320 - {7x\over 2} < 180$ so $x > 40$ and $a < 20$. 
Therefore $\angle DGP = x - a > 20 > a = \angle GDE$. However since $DE = PG$ and $\angle DGP > \angle GDE$, $P$ would have a further distance to $DG$ and $E$ and $PE$ would intersect $GD$ on the extension ray from $G$ to $D$. contradiction.
Case $2$: $EP$ is not parallel to $DG$ and intersect $GD$ on the extension ray from $G$ to $D$. Then this means $\angle GDE + \angle DEP > 180$ which means $320 - {7x\over 2} > 180$ so $x < 40$ and $a > 20$. All similar thing and we can conclude $\angle GDE > \angle DGP$ so $EP$ must intersect $DG$ on the extension from $D$ to $G$. contradiction.
So $EP$ must be parallel to $DG$ and $320 - {7x\over 2} = 180$ and $x=40$.
A: Here is a synthetic solution.
Let $\angle BAC = x$ and $\angle DCE = y$
Now, $\angle ABC = 60 + \frac{y}{3}$. If we prove $y= 30$, the proof would be complete. Let $BC = a$
Note that $$BD = 2a \cos ({60 +\frac{y}{3}})$$
and since $AB= \frac {a}{2\cos ({60 + \frac{y}{3}})}$, we can calculate after simplifying and using the property
 :$\frac {4\cos^2 t -1}{2\cos t} = \frac {3- 4\sin^2 t}{2\cos t}= \frac {\sin 3t}{\sin 2t}$
$$\frac {AD}{a} = \frac {\sin y}{\cos ({30 + \frac {2y}{3}})}$$
By sine rule in $\Delta DEC$, and using the above result,
$$\sin{y}\sin({y+50})= \sin{50}\cos({30+\frac{2y}{3}})$$
And so,
$$2\sin{y}\sin({y+50})= 2\sin{50}\cos({30+\frac{2y}{3}})$$
if $y=30$, then this equation is satisfied because RHS becomes $\sin 100$ and LHS becomes $\sin 80$ and they are obviously equal.
Therefore, $x=40$ 
Note:
To see that that equation has no other solutions than $y=30$, I suggest using Wolfram Alpha.
