Any ideas?

solution idea

Multiply the sides by $2$. They are $|d+e-b-a|$, $|b+c-e-f|$, and $|f+a-c-d|$. The sum of the three things in the absolute value is $0$, so they form a triangle.

Nice idea....

Here is my friend's solution. He told me this in a call while I was swimming in the pool after the TMO21 exam. His solution involved using vectors. Proof: First, consider the vectors, \begin{align*} \overrightarrow{UX}&=\overrightarrow{UB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DX}\\ \overrightarrow{WZ}&=\overrightarrow{WD}+\overrightarrow{DE}+\overrightarrow{EF}+\overrightarrow{FZ}\\ \overrightarrow{YV}&=\overrightarrow{YF}+\overrightarrow{FA}+\overrightarrow{AB}+\overrightarrow{BV} \end{align*}Notice that $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EF}+\overrightarrow{FA}=\overrightarrow{0}$. Which implies \begin{align*} \overrightarrow{UX}+\overrightarrow{WZ}+\overrightarrow{YV}&=\overrightarrow{UB}+\overrightarrow{DX}+\overrightarrow{WD}+\overrightarrow{FZ}+\overrightarrow{YF}+\overrightarrow{BV}\\ &=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EF}+\overrightarrow{FA}\right)=\overrightarrow{0} \end{align*}Therefore, $\overline{UX},\overline{VY},\overline{WZ}$ form a triangle as desired.

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