The problem is equivalent to showing that $Y:=BC\cap AD$ is the center of the circle that passes through $P,Q,R,S$. Using Stewart's, the problem reduces to some side-bashing equations which can be done by Menelaus's.

KereMath wrote: Let $ABCD$ be a quadrilateral such that line $AB$ intersects $CD$ at $X$. Denote circles with inradius $r_1$ and centers $A, B$ as $w_a$ and $w_b$ with inradius $r_2$ and centers $C, D$ as $w_c$ and $w_d$. $w_a$ intersects $w_d$ at $P, Q$. $w_b$ intersects $w_c$ at $R, S$. Prove that if $XA.XB+r_2^2=XC.XD+r_1^2$, then $P,Q,R,S$ are cyclic. Original question has given $XA.XB+r_1^2=XC.XD+r_2^2$, and its sides not parallel. There is a small typo in the post. Let $p$ and $q$ be the radical-axises of the circles $\omega_a, \omega_b$ and $\omega_c,\omega_d$. Let $T$ be their point of intersection. We know that $T$ exists, since $AB$ and $CD$ are not parallel so becuase of $p\perp AB$, $q\perp CD$, $p$ and $q$ are not parallel, too. Now let $q\cap CD=\{E\}$, $p\cap AB=\{F\}$. Given condition simplifies as follows: $$XA.XB+r_1^2=XC.XD+r_2^2\Rightarrow XF^2-AF^2+r_1^2=XE^2-DE^2+r_2^2\Rightarrow (XT^2-TF^2)-AF^2+r_1^2=(XT^2-TE^2)-DE^2+r_2^2\Rightarrow$$$$(-TF^2-AF^2)+r_1^2=(-TE^2-DE^2)+r_2^2\Rightarrow TA^2-r_1^2=TD^2-r_2^2$$ So the power of the point $T$, respective to $\omega_d$ is equal to its power respective to $\omega_a$. Also for any point on $p$, we have its powers relative to $\omega_a$ and $\omega_b$ are equal too. Same goes for $q$, and since the point $T$ is on $p$ and $q$ same time, we conclude that point $T$ is the radical center of these $4$ circles. So $T$ must lie on the radical axis of the circles $\omega_b$ and $\omega_c$ so $T$ lies on $RS$. So $T=PQ\cap RS$. ($PQ$ and $RS$ is not parallel too.) And finally, because $T$'s powers relative to $\omega_a$ and $\omega_b$ is equal, we have $$TQ.TP=TR.TS\Rightarrow P,Q,R,S~\text{are cyclic.}$$

Let $F,G$ be the midpoints of $[AB],[DC]$ respectively. Let $PQ\cap AD$ be $T$. We know that $PQ\perp AD$ so $r_1^2-r_2^2=|AT|^2-|TD|^2$. So $T$ is an only point on $AD$. Let the lines that passes through $F,G$ and perp to $AB,BC$ meets at $H$. If $H$ lies on $PQ$, by radical axis of $w_a,w_b, PQRS$ will be cyclic. Lets prove this. Let $X$'s symmetrys to $F,G$ be $X_1,X_2$ respectively. Let $AD\cap(XX_1D),(XX_2A)$ be $Z_1,Z_2$ respectively. $$pow(A,(XZ_1D))=|Z_1A||AD|=|XA||AX_1|=|XA||XB|,pow(D,(AXZ_2))=|AD||DZ_2|=|XD||DX_2|=|XC||XD|$$so $|XC||XD|-|XA||XB|=|AD|(|Z_2D|-|Z_1A|)$ so we have to prove $|AT|^2-|TD|^2=|AD|(|Z_2D|-|Z_1A|)=|AD|(|AT|-|TD|)$ $\Longleftrightarrow$ we have to prove $T$ is the midpoint of $[Z_1Z_2]$. Let $Z_1Z_2$ meets on $J$. By angle chasing we can say $JX_1HX_2$ is cyclic. And its clear that $|HX_1|=|HX_2|$ so $JH$ is angle bisector of $\angle{Z_1JZ_2}$. We know that $Z_1Z_2J$ is isosceles triangle so $J,H,T$ is collinear. $T$ is midpoint. We're done.