geometry
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
$\triangle ABC$ is isosceles. Therefore,
$\measuredangle BAC = \measuredangle C \implies 2\alpha+\alpha =120^\circ\implies \alpha =40^\circ$
Therefore $\measuredangle ABC = 180^\circ - 160^\circ = 20^\circ$
Draw $AX$ such that $\measuredangle XAK = 30^\circ$.
Draw $TX$.
$\measuredangle CDB = 120^\circ ~~\therefore \measuredangle DIA = 90^\circ$
Since $CI$ is angle bisector $IA = IX$
$\triangle ITA \cong \triangle IXT ~(L.A.L) \rightarrow \measuredangle AXT = 60^\circ$
Therefore $\triangle ATX$ is equilateral.
$\measuredangle KXB = 180 ^\circ -110^\circ= 70^\circ$
$AK$ is angle bisector and the height of triangle $ATX$
$ \therefore AKT = 90^\circ$
$TK = KX \rightarrow \triangle KTB \cong \triangle KXB~ (L.A.L)$
Therefore $\triangle TBX$ is isosceles.
$\measuredangle B = 180^\circ -140^\circ =40^\circ$
$\therefore \boxed{\color{red}x=40^\circ-20^\circ=20^\circ}$
$\widehat {ECG}=\frac{(ECF=90+\alpha)}2=45+\alpha$
$\widehat {ECA}=\widehat {ECG}=45+\frac{\alpha}2$
$\widehat {EAC}=90-(45+\frac{\alpha}2)=45-\frac{\alpha}2$
Therefore:
$\widehat {ECG}=\widehat {AEG}$
that is triangles ECA and EGA are similar, since $\widehat {CEA}=90$, therefore :
$\widehat {EGA}=90^o$
Best Answer
$ABCD$ 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 $\angle ABC \lt 60^\circ$.