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Let's assume that $\triangle ABC$, $\triangle ADE$ are equilateral triangles, and that $|DC| = 1, |EC| = 2$. How could we find $|DE|$?

Since $\triangle ABC$, $\triangle ADE$ are equilateral triangles, if $\angle BAD = \alpha$, $\angle DAC = \beta$, and $\angle GAE = \theta$, then $\alpha+\beta = 60^{\circ}$ and $\beta + \theta = 60^{\circ}$, from which we conclude that $\alpha = \theta$. And since $\angle DCG = 60^{\circ}$,

$$|DE| = \sqrt{5+4\cos(60^{\circ}+\angle GCE)}$$

But for $\angle GCE$ to be determined, we'll have to determine $\angle GDC$ and $\angle GEC$.

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  • $\begingroup$ Where is point $G$? $\endgroup$
    – peterwhy
    Jan 19, 2023 at 14:25
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    $\begingroup$ As hinted in Sathvik's answer, $AECD$ is a cyclic quadrilateral. Then these opposite angles add to $180^\circ$: $\angle DCE + \angle DAE = 180^\circ$ or $\angle DCE + 60^\circ = 180^\circ$. $\endgroup$
    – peterwhy
    Jan 19, 2023 at 14:45

1 Answer 1

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Hint:

$\angle ACB=\angle AED=60^{\circ}\implies AECD$ is a cyclic quadrilateral.

Using Ptolemy's Theorem, $|AC|=|EC|+|DC|=3$. Apply the cosine rule in $\triangle ABD$ to find $|AD|=|DE|$.

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