Find $\angle x$ in the figure if ABCD is a rhombus, BEC is equilateral and For reference: Find $\angle~ x$ in the figure if ABCD is a rhombus, BEC is equilateral and $\measuredangle BCT = 2m\measuredangle BAE$ (answer: $60^\circ$)
original figure:

My progress:
$ABCD : square\\
ABC: \triangle equilateral \implies \theta = 30^\circ\\
\measuredangle BET = 60^\circ$
still missing an equation

I couldn't draw with the "rhombus" itself, only with the particular case of a square that is a rhombus
 A: 
$ABCD$ is a rhombus and $\triangle BCE$ is equilateral.
Also, $\angle BCT = 2 \angle BAE$.
Please note position of point $T$ on $AE$ changes as the angles of rhombus change. When $D, C, E$ are collinear (the acute angle of rhombus being $60^\circ$), $T$ falls on vertex $A$. I have shown above the construct for acute angle of rhombus being greater than $60^\circ$. What you drew (a square) is a specific case. When acute angle of rhombus is less than $60^\circ$, point $T$ is outside rhombus on $EA$ extend such that $\angle BCT = 2 \angle BAE$. See a diagram showing this construct at the end of the answer.

Now coming to the solution,
Say, $\angle ABC = \theta$ then $\angle ABE = 60^\circ + \theta$. $\angle BAE = \angle BEA = 60^\circ - \frac{\theta}{2}, \angle AEC = \frac{\theta}{2}$
$\angle BCT = 2 \angle BAE = 120^\circ - \theta, \angle ECT = 180^\circ - \theta$
So, $\angle CTE = \frac{\theta}{2}$ and $\triangle ECT$ is isosceles and $CT = EC = CD$.
$\angle DCT = \angle BCD - \angle BCT = 60^\circ$
That concludes $\triangle DCT$ is equilateral and hence, $\angle CTD = 60^\circ$

A construct with $\angle BCT = 2 \angle BAE$ and $\angle ABC \lt 60^\circ$.

A: 
Hints: Triangle ABE is isosceles so:
$\angle BAE=\angle BEA=\frac{\angle BCT}2$
that is T is on the circle center at C, so $TC=BC=CD$
Now if you proove $TD=TC$ then $x=60^o$
