# 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

• 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:". Commented Aug 22, 2021 at 14:06
• @coffeemath...thanks for the alert..I've corrected it Commented Aug 22, 2021 at 14:13
• $CE=DE=ET$ and $\angle TED=60$, so $\triangle ETD$ is equilateral Commented Aug 22, 2021 at 14:20
• 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). Commented Aug 22, 2021 at 15:26
• @coffeemath see now Commented Aug 22, 2021 at 15:55

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

• @Math..excellent grateful Commented Aug 22, 2021 at 19:22
• You are welcome! Commented Aug 22, 2021 at 23:20

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 $$x=60^o$$