3. Lemma. If $U, W, V$ are three points on a line $l$ in this order, and $X$ a point in the plane with $X W \perp U V$, then $\angle U X V = 90^{\circ}$. Proof. Let $X W^{2} > U W \cdot V W$, and let $X_{0}$ be a point on the segment $X W$ such that $X_{0} W^{2} = U W \cdot V W$. Then $X_{0} W / U W = V W / X_{0} W$, so that triangles $X_{0} W U$ and $V W X_{0}$ are similar. Thus $\angle U X_{0} V = \angle U X_{0} W + \angle W U X_{0} = 90^{\circ}$, which immediately implies that $\angle U X V = 90^{\circ}$. Note that $\triangle B K R \sim \triangle B S L$ : in fact, we have $\angle K B R = \angle S B L = 90^{\circ} - \beta / 2$ and $\angle B K R = \angle A K M = \angle K L M = \angle B S L = 90^{\circ} - \alpha / 2$. In particular, we obtain $B R / B K = B L / B S = B K / B S$, so that $B R \cdot B S = B K^2$. Since $B M > B L$, we conclude that $\angle M B E < \frac{1}{2} \angle M B K$ and $\angle M B F < \frac{1}{2} \angle M B L$. Adding these two inequalities gives $\angle E B F < \beta / 2$. Therefore $\angle R I S < 90^{\circ}$. Remark. It can be shown (using vectors) that the statement remains true for an arbitrary line $t$ passing through $B$.