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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: enter image description here

My progress:

$ABCD : square\\ ABC: \triangle equilateral \implies \theta = 30^\circ\\ \measuredangle BET = 60^\circ$

still missing an equation enter image description here

I couldn't draw with the "rhombus" itself, only with the particular case of a square that is a rhombus

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  • $\begingroup$ It seems you changed some labels from original diagram. You labelled the old C as E and the old E as C. This should be fixed in your new diagram, and in the section "My progress:". $\endgroup$
    – coffeemath
    Aug 22, 2021 at 14:06
  • $\begingroup$ @coffeemath...thanks for the alert..I've corrected it $\endgroup$ Aug 22, 2021 at 14:13
  • $\begingroup$ $CE=DE=ET$ and $\angle TED=60$, so $\triangle ETD$ is equilateral $\endgroup$ Aug 22, 2021 at 14:20
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    $\begingroup$ Depending on whether $CD$ and $CE$ are collinear or not, the figure could either be an isosceles trapezoid (as it appears to be in the first figure) or a pentagon (as it appears to be in the second figure). $\endgroup$ Aug 22, 2021 at 15:26
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    $\begingroup$ @coffeemath see now $\endgroup$ Aug 22, 2021 at 15:55

2 Answers 2

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enter image description here

$ABCD$ 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 $\angle ABC \lt 60^\circ$.

enter image description here

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    $\begingroup$ @Math..excellent grateful $\endgroup$ Aug 22, 2021 at 19:22
  • $\begingroup$ You are welcome! $\endgroup$
    – Math Lover
    Aug 22, 2021 at 23:20
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enter image description here

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 $x=60^o$

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