3. Let points $A, B, C$ be collinear in this order, and the rays $(B D$ and $(B E$ be in the same half-plane determined by the line $A C$ such that $(B D$ and $(B E$ are perpendicular. What values can the measure of the angle formed by the bisectors of angles $\varangle A B E$ and $\varangle D B C$ take?
## Solution and Scoring:

Let ( $B M$ and ( $B N$ be the bisectors of angles $\varangle A B E$ and $\varangle D B C$, respectively.

We distinguish two cases: I. $(B E \in \operatorname{Int}(\varangle A B D)$ or II. $(B D \in \operatorname{Int}(\varangle A B E) \ldots . \mathbf{2 p}$ Case I. If we denote $m(\varangle A B E)=a$ and $m(\varangle C B D)=b$, we have $a+90^{\circ}+b=180^{\circ}$, thus $a+b=90^{\circ}$. Then $m(\varangle M B N)=m(\varangle M B E)+m(\varangle E B D)+m(\varangle D B N)=$ $\frac{a}{2}+90^{\circ}+\frac{b}{2}=135^{\circ}$.

Case II. With the same notations, we have $a+b-90^{\circ}=180^{\circ}$, thus $a+b=270^{\circ}$. We have that $\frac{a}{2}+\frac{b}{2}=m(\varangle D B N)+m(\varangle M B E)=m(\varangle D B E)+m(\varangle M B N)$, from which $m(\varangle M B N)=\frac{a}{2}+\frac{b}{2}-90^{\circ}=45^{\circ}$.

$3 p$