In the triangle $ABC$ it is $AC = BC$ and $\alpha = \beta$. The points $D$ and $E$ are on the line through $A$ and $B$. Show that the triangle $CDE$ is isosceles.
Hey there! Is it sufficient to say that:
$$\triangle_{CDE} \text{ isoceles } \Longleftrightarrow \measuredangle(EDC) = \measuredangle(CED)$$
because ($\gamma := \measuredangle(DAC)$ and $\delta := \measuredangle(CBE)$)
$$*\quad AC = BC \Longrightarrow \gamma = \delta$$ $$*\quad \measuredangle(ADC) = 180° - \alpha - \gamma$$ $$\quad \measuredangle(BEC) = 180° - \beta - \delta$$ $$*\quad \measuredangle(EDC) = 180° - (180° - \alpha - \gamma) = \alpha + \gamma$$ $$\quad \measuredangle(CED) = 180° - (180° - \beta - \delta) = \beta + \delta$$ $$* \quad \measuredangle(EDC) = \alpha + \gamma = \beta + \gamma = \measuredangle(CED) \Longrightarrow \triangle_{CDE} \text{ isoceles }$$
As dxiv said, Using exterior angle property will be shortest approach here.
$$\angle A + \alpha=\angle B+\beta$$ $$\therefore\angle CDE = \angle CED$$
Another approach can be proving congruency of $\triangle CAD$ and $\triangle CBE$ by $ASA$ congruency:
$$CA = CB\ , \\ \alpha=\beta\ \\ \angle A = \angle B$$
$$\triangle CAD \cong\triangle CBE \ \ \ \implies \ CD=CE$$