Let $r$ be the line passing through $A$ and $B$, and let $s$ be the line passing through $C$ and $D$. Let $M=PQ\cap BC$ and $N=PQ\cap AD$. Due to the definition of point $P$, then $P$ equidists from lines $r$, $BC$ and $s$. Similarly, point $Q$ equidists from lines $r$, $AD$ and $s$. Therefore, both points $P$ and $Q$ equidist from the parallel lines $r$ and $s$ , and so $M$ and $N$ are the midpoints of segments $BC$ and $AD$, respectively. As a result, $$ MN=\frac{AB+CD}{2} \quad (1) $$Now, since $r\parallel s$, then $\angle CBA+DCB=180^o$, and so $\angle PBC+\angle BCP=90^o$. This means that triangle $\triangle BCP$ is rectangle in $P$, so $$ PM=MB=MC \Longrightarrow MP=\frac{BC}{2}\quad(2) $$ Analogously, $\angle BAD+\angle ADC=180^o$ implies $\angle QDA+\angle DAQ=90^o$. This means that triangle $\triangle ADQ$ is rectangle in $Q$, and so $$ NQ=AN=ND \Longrightarrow NQ=\frac{AD}{2} \quad(3) $$Finally, from $(1), (2)$ and $(3)$ one gets $$ PQ=PM+MN+NQ=\frac{AB+BC+CD+DA}{2} $$The problem has been solved.

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