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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 $\delta := \measuredangle(CBE)$)

$$*\quad AC = BC \Longrightarrow \gamma = \delta$$ $$*\quad \measuredangle(ADC) = 180° - \alpha - \gamma$$ $$\quad \measuredangle(BEC) = 180° - \beta - \delta$$ $$*\quad \measuredangle(EDC) = 180° - (180° - \alpha - \gamma) = \alpha + \gamma$$ $$\quad \measuredangle(CED) = 180° - (180° - \beta - \delta) = \beta + \delta$$ $$* \quad \measuredangle(EDC) = \alpha + \gamma = \beta + \gamma = \measuredangle(CED) \Longrightarrow \triangle_{CDE} \text{ isoceles }$$

Thanks!

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  • $\begingroup$ Your proof is correct. You could maybe save a couple of steps by deriving $\angle EDC = \angle DAC + \angle ACD = \gamma + \alpha$ directly from the exterior angle theorem. $\endgroup$ – dxiv Feb 1 '16 at 23:40
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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:

$$CA = CB\ , \\ \alpha=\beta\ \\ \angle A = \angle B$$

$$\triangle CAD \cong\triangle CBE \ \ \ \implies \ CD=CE$$

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