$ \tan \frac{\angle DCE}{2} = \frac{\tan \frac{\pi}{3}}{3} = \tan \frac{\pi}{6}$ then $ \angle DCE = \frac{\pi}{3}$ and $ \triangle CDE$ is equilateral.

See that AD*AB = AC^2, hence <ACD = <ABC = 30, consequently AD = CD. Similarly CE = BE, but AD = DE = BE, so the triangle CDE is eqilateral. Best regards, sunken rock

STARS wrote: Triangle $ ABC$ is isosceles with $ AC = BC$ and $ \angle{C} = 120^o$. Points $ D$ and $ E$ are chosen on segment $ AB$ so that $ |AD| = |DE| = |EB|$. Find the sizes of the angles of triangle $ CDE$. A simple Solution If $ M$ is the midpoint of $ AB$ and $ K,L$ are the projections of $ D,E$ to $ AC,BC$ respectively , then $ DK = \frac {1}{2} AD = DM$. Thus $ DK = DM$ , so $ CD$ is the bisector of $ \angle ACM = 60$. From now on the solution can be done easilly. Babis Note I had proposed this problem in this forum some months ago !

Call $ \widehat{CDE}=\widehat{CED}=\alpha$, $ \widehat{DCE}=\beta$ and $ \widehat{BCE}=\widehat{ACD}=\gamma$ You have $ \beta + 2\alpha = \pi$ and $ \beta + 2\alpha = \pi \Rightarrow \alpha = \gamma + 30$ Apply the sin theorem on $ \Delta CDE$ $ \Rightarrow \frac{CE}{\sin \alpha}=\frac{DE}{\sin \beta} \Rightarrow \frac{\sin 2\alpha}{\sin \alpha}=\frac{DE}{CE}$ and on $ \Delta BCE$ $ \Rightarrow \frac{CE}{\sin 30}=\frac{EB}{\sin \gamma} \Rightarrow \frac{\sin (\alpha-30)}{\sin 30}=\frac{DE}{CE}$ So $ \frac{\sin (\alpha-30)}{\sin 30}=\frac{\sin 2\alpha}{\sin \alpha} \Rightarrow$ $ \sin\alpha\cos30 - \cos\alpha\sin30 = \cos\alpha$ $ \Rightarrow \frac{\sqrt 3}{2}\sin\alpha = \frac{3}{2}\cos\alpha \Rightarrow$ $ \cos^2 \alpha = \frac{1}{4} \Rightarrow \alpha = 60$