1. Identify the properties of the quadrilateral:
Given quadrilateral $ABCD$ with $AB \perp CD$ and $AD \perp BC$, we can infer that $B$ is the orthocenter of $\triangle ACD$.

2. Lemma 1:
In any $\triangle XYZ$ and a point $P$ in the plane, let $PD$ and $PE$ be perpendicular lines from $P$ to $YZ$ and $XZ$ respectively. If $\frac{PD}{YZ} = \frac{PE}{XZ}$, then $P$ lies on the $Z$-symmedian of $\triangle XYZ$.

Proof of Lemma 1:
- Let $ZP$ intersect $\odot(XYZ)$ at $W$.
- Using the sine rule in $\triangle PZY$ and $\triangle PZX$:
$$\frac{PD}{PE} = \frac{\frac{PD}{PZ}}{\frac{PE}{PZ}} = \frac{\sin \angle PZY}{\sin \angle PZX} = \frac{\sin \angle WZY}{\sin \angle WZX}$$
- Since $\angle WZY$ and $\angle WZX$ are angles subtended by the same arc in the circumcircle, we have:
$$\frac{\sin \angle WZY}{\sin \angle WZX} = \frac{\sin \angle WXY}{\sin \angle WYX} = \frac{WY}{WX}$$
- Therefore, $\frac{YZ}{XZ} = \frac{WY}{WX}$, implying $WXZY$ is a harmonic quadrilateral.
- Hence, $P$ lies on the $Z$-symmedian of $\triangle XYZ$.

3. Application to the original problem:
- Let $j, k$ be the tangents to $\odot(ABD)$ at $B, D$ respectively, and let $l, m$ be the tangents to $\odot(BCD)$ at $B, D$ respectively.
- Let $CF$ and $AE$ be the $C$-symmedian of $\triangle BCD$ and $A$-symmedian of $\triangle ABD$ respectively, with $j \cap k = E$ and $l \cap m = F$. Let $AE \cap CF = P$.

4. Claim:
$P$ is the required point such that the distances from $P$ to the sidelines are proportional to the lengths of the corresponding sides.

5. Proof of the claim:
- Since $P$ lies on the $C$-symmedian of $\triangle BCD$ and the $A$-symmedian of $\triangle ABD$, we have:
$$\frac{PD}{BC} = \frac{PG}{BC} \quad \text{and} \quad \frac{PE}{AD} = \frac{PF}{AB}$$
- To prove $\frac{PD}{BC} = \frac{PE}{AD}$, we need to show that $DG, AE, CF$ are concurrent.

6. Concurrent lines:
- Let $n, o$ be the tangents to $\odot ACD$ through $A, C$ respectively, and let $n \cap o = G$.
- We need to prove $DG, AE, CF$ are concurrent.
- Using Desargues' Theorem on $\triangle ACG$ and $\triangle EFD$:
- Since $AC \parallel EF$, $CG \parallel FD$, and $AG \parallel ED$, the three pairs of lines are parallel.
- Therefore, the intersection points $AC \cap EF$, $CG \cap FD$, and $AG \cap ED$ are collinear (on the line at infinity).

7. Conclusion:
- Thus, $AE, CF$, and $GD$ are concurrent.
- Therefore, there exists a unique point $P$ such that the distances from $P$ to the sides of the quadrilateral are proportional to the lengths of the sides respectively.

$\blacksquare$