One solution would say both sides are equivalent with a sum of vectors is zero, someone can explain?

Consider vectors $a= \vec{AB},b= \vec{BC},...,f= \vec{FA}$. First, we will show: $\frac{PQ}{CD} = \frac{QR}{EF} = \frac{RP}{AB} \Longleftrightarrow a+c+e=0$. Proof: First, the forward implication:$\frac{PQ}{CD} = \frac{QR}{EF} = \frac{RP}{AB}$ implies that we can scale down vectors $ \vec{PQ}, \vec{QR}, \vec{RP}$ by the same amount, to produce vectors $a,c,e$. Since $\vec{PQ}+ \vec{QR}+ \vec{RP}=0$, we have $a+c+e=0$. Now, the reverse implication. Suppose that $a+c+e=0$. Consider the triangle formed by lining $a,c,e$ head to tail (It exists since they sum to 0). Now, this triangle has sides parallel to $PQR$, so they are similar. Thus, the sides have the same ratios of lengths: $\frac{PQ}{CD} = \frac{QR}{EF} = \frac{RP}{AB}$. Clearly, we have $a+b+c+d+e+f=0$, since the vectors form a closed path. As a result, we have $a+c+e=0 \Longleftrightarrow b+d+f=0$. Since $\frac{PQ}{CD} = \frac{QR}{EF} = \frac{RP}{AB} \Longleftrightarrow a+c+e=0$, and $\frac{ST}{DE} = \frac{TU}{FA} = \frac{US}{BC} \Longleftrightarrow b+d+f=0$, the two given length relations are equivalent.

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