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Let $ABC$ and $CDE$ be equilateral triangles.

How to prove that $x=120^\circ$?

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Thank you.

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**Hint:**

- Observe that $\triangle ACE$ is $\triangle BCD$ rotated by $60^\circ$ degrees.
- The above implies that $|\angle AXB| = 60^\circ$, where $X$ is the intersection of $AE$ and $BD$.

$\hspace{50pt}$

I hope this helps $\ddot\smile$

**NOTE**: This proof assumes that the segment $AE$ and $BD$ actually intersect at point $F$. Note that the $x=120^{\circ}$ holds when the the lines $AE$ and $BD$, instead of the segments intersect, but you can prove it in a simular fashion.

Let $F$ be the intersection point of $AE$ and $BD$ be $F$. Then from the quadrlaterial $FACD$ we have:

$$\angle AFD + \angle FDC + \angle DCA + \angle CBA = 360^{\circ}$$

$$120^{\circ} + 60^{\circ} – \angle EDF + 60^{\circ} + 60^{\circ} + \angle BCE + 60^{\circ} – \angle BAF = 360^{\circ}$$

$$\angle BCE = \angle BAF + \angle EDF \tag{1}$$

Now for the $\triangle ACE$ we have:

$$\angle EAC + \angle ACE + \angle CEA = 180^{\circ}$$

$$60^{\circ} – \angle BAF + 60^{\circ} + \angle BCE + \angle CEA = 180^{\circ}$$

Nos using $(1)$ we have:

$$\angle CEA = 60^{\circ} – \angle EDF$$

Now using this we have:

$$\angle FED = 60^{\circ} + \angle CEA = 120^{\circ} – \angle EDF$$

Now using this from the $\triangle FED$ we have:

$$\angle DFE = 180^{\circ} – 120^{\circ} + \angle EDF – \angle EDF$$

$$\angle DFE = 60^{\circ}$$

Now at last:

$$\angle AFD = 180^{\circ} – \angle DFE = 180^{\circ} – 60^{\circ} = 120^{\circ}$$

Q.E.D.

Let intersection of AE and BD be F. G is a point on AE, such that CG is parallel to BD. $\angle CBD = \angle BCG =\beta \Rightarrow \angle GCA =60^\circ – \beta $

Triangles ACE and BCD are equal, therefore $ \angle EAC = \beta $

Therefore $ \angle CGA =120^\circ $

Triangles ACE and BCD are equal, because two sides and angle between them are equal.

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