# Determining $|DE|$

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 $$\angle GEC$$.

• Where is point $G$? Jan 19, 2023 at 14:25
• 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 $\angle DCE + 60^\circ = 180^\circ$. Jan 19, 2023 at 14:45

$$\angle ACB=\angle AED=60^{\circ}\implies AECD$$ is a cyclic quadrilateral.
Using Ptolemy's Theorem, $$|AC|=|EC|+|DC|=3$$. Apply the cosine rule in $$\triangle ABD$$ to find $$|AD|=|DE|$$.