On the $BE$ side of a regular $ABE$ triangle, a $BCDE$ rhombus is built outside it. The segments $AC$ and $BD$ intersect at point $F$. Prove that $AF <BD$.
Firstly notice that $|EA|=|EB|=|ED|$. Hence, $E$ is the circumcenter of $ABD$ and $C$ is the reflection of $E$ across $BD$. Let $X$ be the reflection of $A$ across $BD$. Since the rhombus was constructed outside of $ABE$ we have that $\measuredangle{ADE}=\frac{\measuredangle{AEB}}{2}=\frac{\pi}{6}$. Otherwise the angle would have equaled $\frac{5\pi}{6}$. Because $|EA|=|ED|$ we have $|CD|=|CX|$. To prove that $A$ lies on the perpendicular bisector of $DX$ as well note that $\measuredangle{XAD}=\frac{\pi}{2}-\measuredangle{ADB}=\frac{\pi}{3} \Rightarrow AXD$ is equilateral. Thus $AC\perp DX\Rightarrow \measuredangle{CAD}=\frac{\pi}{2}-\measuredangle{ADX}=\frac{\pi}{6}$. Let $F=DB\cap AC$. Now it is evident that $\measuredangle{FAD}=\frac{\pi}{6}=\measuredangle{ADF}$ implying that $AFD$ is isoceles. Since $F$ is on the segment $BD$ we have both $|DF|>0$ and $|FB|> 0$. To conclude notice that $|DB|=|DF|+|FB|>|DF|=|AF|$.