Problem 1598

$\angle B$ can be $22.5^\circ$ or $67.5^\circ$.

WLOG assume $AC < BC$. Let $D$ be the foot of altitude from $C$, let $E$ be the foot of internal angle bisector from $C$, and let $F$ be the midpoint of $AB$. Let $P$ be the projection of $E$ on $BC$, and let $DP$ meet $AC$ at $Q$.

Firstly, since $\angle ACD = \angle ECD$ and $CD \perp AE$, we have $\triangle ACD \cong \triangle ECD$. Secondly, note that $C, D, E, P$ are concyclic since
$$\angle CDE + \angle CPE = 90^\circ + 90^\circ = 180^\circ.$$

$$\angle CPQ = \angle CPD = \angle CED = \angle CAD = \angle CAB.$$

Together with the common angle at $C$, we know that $\triangle CPQ \sim \triangle CAB$ (and $D$ lies between $P$ and $Q$). As $\angle DCQ = \angle FCB$, the points $D$ and $F$ are corresponding points under this similarity. Thus, $D$ is the midpoint of $PQ$. Now, since $D$ is also the midpoint of $AE$, the quadrilateral $AQEP$ is a parallelogram. This implies $CQ // PE$, and hence $\angle ACB = 90^\circ$. Then we have
$$\angle ABC = \angle FCB = \frac{90^\circ}{4} = 22.5^\circ.$$
If $AC > BC$, then the roles of $A$ and $B$ are swapped, and hence $\angle ABC = 67.5^\circ$.