Geometry with triangle.

Question

Any other solutions(advice) are welcome.

$\angle BAC=60^\circ, \;\;\;\angle ACB=x,\;\;\; \overline {BD}=\overline{BC}=\overline{CE} $

Answer 1

$$\angle CBA=120^\circ-x$$

Triangle $DBC$ is isosceles:

$$\angle CDB=\angle DCB=\frac12\angle CBA=60^\circ-\frac x2$$

Triangle $BCE$ is isosceles:

$$\angle CEB=\angle CBE=\frac12\angle BCA=\frac x2$$

Now, from triangle:

$$\angle BFC=120^\circ$$

Take a look at quadrilateral $ABFC$: The sum of opposite angles $\angle A$ and $\angle F$ is $180^\circ$ so quadrialteral $ABCD$ is cyclic. Because of that:

$$\angle FAB=\angle FCB=60^\circ-\frac x2=\angle FDB$$

So triangle $FAD$ is isosceles and:

$$FA=FD\tag{1}$$

In a similar way:

$$\angle FAC=\angle FBC=\frac x2=\angle FEA$$

So triangle $FEA$ is isosceles and:

$$FA=FE\tag{2}$$

From (1) and (2) you have:

$$FD=FE$$

...and triangle $FDE$ must be isosceles. Angle $\angle DFE=120^\circ$ and therefore:

$$\angle FDE=\angle FED=30^\circ$$

It follows that:

$$\angle DEA=\angle DEF+\angle FEA=30^\circ+\frac x2$$

Answer 2

Drawing the rhombus is a nice way to start but we can finish differently.

As soon as you find that $\triangle BDF$ is equilateral, you know that $FD = FB = FE.$ That is, the points $D,$ $B,$ and $E$ all lie on the same circle with center at $F.$

Referring to this same circle, since the central angle $\angle BFD$ is $60^\circ,$ the inscribed angle $\angle BED$ is $30^\circ.$

We have $\angle CEF = \angle ACB = x$ and $EB$ is the angle bisector of $\angle CEF,$ so $\angle CEB = \frac12 \angle CEF = \frac 12x.$

Then $\angle CED = \angle CEB + \angle BED = \frac 12x + 30^\circ.$