10.1.6 ** Given a square $A B C D$, with points $S, P$ on $A B, B C$ respectively, and $D S, D P$ intersecting $A C$ at points $R, Q$, find the necessary and sufficient condition for points $B, P, Q, R, S$ to be concyclic (i.e., what position relationship must points $S, P$ satisfy).
Parse the figure, let's assume the side length of the square is $1, AS=a, CP=b$. We will prove that the necessary and sufficient condition is $BS \cdot BP=2 AS \cdot CP$. First, we prove the sufficiency. Let $\angle ASD=\alpha, \angle DPC=\beta$, then $\tan (\alpha+\beta)=$ $\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{\frac{1}{a}+\frac{1}{b}}{1-\frac{1}{ab}}=\frac{a+b}{ab-1}$. Since $(1-a)(1-b)=2ab$, we have $\tan (\alpha+\beta)=-1, \alpha+\beta=135^{\circ}$.

Connect $BD, BR$, it is easy to see that $\triangle BRC \cong \triangle DRC$, so $\angle RBC=\angle RDC=\angle ASD=\alpha$. And $\angle RQP=\beta+45^{\circ}$, hence $\angle RBC+\angle RQP=\alpha+\beta+45^{\circ}=180^{\circ}$, which means points $B, R, Q, P$ are concyclic. Similarly, points $B, S, R, Q$ are concyclic. Therefore, $B, S, R, Q, P$ are five concyclic points.

Conversely, if $B, S, R, Q, P$ are five concyclic points, we can get $\alpha+\beta=\angle RQD+\beta=135^{\circ}$, thus $BS \cdot BP=2 AS \cdot CP$. In summary, the necessary and sufficient condition is $BS \cdot BP=2 AS \cdot CP$.