Let $d(P, l)$ denote the distance from point $P$ to line $l$. We first prove the following lemma.

Lemma Let $\angle SAT = \alpha$ be a given angle and $P$ a moving point in $\angle SAT$. If the sum of distances from $P$ to lines $AS$ and $AT$ is a constant number $m$, then the trace of $P$ is a segment of $BC$ with points $B$ and $C$ on $AS$ and $AT$ respectively, and $AB = AC = \frac{m}{\sin \alpha}$. Further, if a point $Q$ lies inside $\triangle ABC$, then the sum of distances from $Q$ to lines $AS$ and $AT$ is less than $m$, and it is greater than $m$ if $Q$ lies outside $\triangle ABC$.

Proof of Lemma For $AB = AC = \frac{m}{\sin \alpha}$ and $P$ on $BC$, we have $S_{\triangle PAB} + S_{\triangle PAC} = S_{\triangle ABC}$, i.e. $\frac{1}{2}AB \cdot d(P, AB) + \frac{1}{2}AC \cdot d(P, AC) = \frac{1}{2} \cdot AB \cdot AC \cdot \sin \alpha$. Then $d(P, AB) + d(P, AC) = m$. If point $Q$ lies inside $\triangle ABC$, $S_{\triangle QAB} + S_{\triangle QAC} < S_{\triangle ABC}$, then $d(Q, AB) + d(Q, AC) < m$. If $Q$ lies outside $\triangle ABC$, $S_{\triangle QAB} + S_{\triangle QAC} > S_{\triangle ABC}$, then $d(Q, AB) + d(Q, AC) > m$. The proof of Lemma is complete.

We now consider the following two cases:

(1) Neither pair of the opposite sides of the quadrilateral $ABCD$ is parallel. We may assume that sides $BC$ and $AD$ meet at point $F$ and sides $BA$ and $CD$ meet at point $E$. Through point $P$ draw segments $l_1$ and $l_2$ such that the sum of distances from any point on $l_1$ to $AB$ and $CD$ is constant and the sum of distances from any point on $l_2$ to $BC$ and $AD$ is also constant. As seen in the second figure, for any point $Q$ in area $S$, using the given condition and the lemma, we have

$$\begin{align*} d(P, AB) + d(P, BC) + d(P, CD) + d(P, DA) \\ &= d(Q, AB) + d(Q, BC) + d(Q, CD) + d(Q, DA) \\ &= [d(Q, AB) + d(Q, CD)] + [d(Q, BC) + d(Q, DA)] \\ &> [d(P, AB) + d(P, CD)] + [d(P, BC) + d(P, DA)]. \end{align*}$$

It leads to a contradiction.

(2) The quadrilateral $ABCD$ is a trapezoid. In the same way, we can also show that it will lead to a contradiction.

This completes the proof.