Applying Menelaus' theorem (孟氏定理) to $\triangle ABC$ with the line $PQ$, and to $\triangle ACD$ with the line $RS$, we see that the necessary and sufficient condition for $AC$ to meet $PQ$ and $RS$ at the same point (this point may be at infinity) respectively is:
$$\frac{AP}{PB} \cdot \frac{BQ}{QC} \cdot \frac{CR}{RD} \cdot \frac{DS}{SA} = 1. \qquad (1)$$
So we set our goal to proving equation (1).

First let us prove the following result.

Lemma 1. Suppose the quadrilateral $EFGH$ has an incircle with center $M$. Then
$$\frac{EF \cdot FG}{GH \cdot HE} = \frac{FM^2}{HM^2}.$$

Proof. Note that $\angle EMH + \angle GMF = \angle FME + \angle HMG = 180^\circ$, $\angle FGM = \angle MGH$, and $\angle HEM = \angle MEF$ (as in Figure 1). By the Law of Sines, we know
$$\begin{aligned} \frac{EF}{FM} \cdot \frac{FG}{FM} &= \frac{\sin \angle FME \cdot \sin \angle GMF}{\sin \angle MEF \cdot \sin \angle FGM} = \frac{\sin \angle HMG \cdot \sin \angle EMH}{\sin \angle MGH \cdot \sin \angle HEM} \\ &= \frac{GH}{HM} \cdot \frac{HE}{HM}. \end{aligned}$$

Let the points $I, J, K, L$ be the incircle centers of the quadrilaterals $APOS, BQOP, CROQ, DSOR$ respectively. Applying the result of Lemma 1 to these four quadrilaterals, we obtain
$$\frac{AP \cdot PO}{OS \cdot SA} \cdot \frac{BQ \cdot QO}{OP \cdot PB} \cdot \frac{CR \cdot RO}{OQ \cdot QC} \cdot \frac{DS \cdot SO}{OR \cdot RD} = \frac{PI^2}{SI^2} \cdot \frac{QJ^2}{PJ^2} \cdot \frac{RK^2}{QK^2} \cdot \frac{SL^2}{RL^2},$$
which simplifies to
$$\frac{AP}{PB} \cdot \frac{BQ}{QC} \cdot \frac{CR}{RD} \cdot \frac{DS}{SA} = \frac{PI^2}{PJ^2} \cdot \frac{QJ^2}{QK^2} \cdot \frac{RK^2}{RL^2} \cdot \frac{SL^2}{SI^2}. \quad (2)$$

Figure 2

Next, we have $\angle IPJ = \angle JOI = 90^\circ$, and the points $I, J$ lie on opposite sides of the line $OP$. From this we know that the quadrilateral $IPJO$ has a circumscribed circle. Similarly, the quadrilateral $JQKO$ also has a circumscribed circle, with $\angle JQK = 90^\circ$. Therefore $\angle QKJ = \angle QOJ = \angle JOP = \angle JIP$.

Hence the two right triangles $\triangle IPJ$ and $\triangle KQJ$ are similar, from which we know $\frac{PI}{PJ} = \frac{QK}{QJ}$. Similarly we obtain $\frac{RK}{RL} = \frac{SI}{SL}$. Combining these two equations with equation (2), we get equation (1). This completes the proof.