Problem 1259

Denote by $I$ the incenter of the triangle $\triangle ABC$ (fig. 03). Also denote the point $D_1$, which is the symmetric reflection of $D$ with respect to $BE$. Then $DB = BD_1$ and $DI = ID_1$.

Without any dependence on the given conditions we have $D_1 \in AB$, since in $\triangle DBD_1$ the line $BE$ contains the altitude, so it is the bisector of $\angle DBD_1$.

If $\angle ACB = 60^\circ$ we get $\angle AIB = 90^\circ + \frac{1}{2}\angle ACB = 120^\circ$. Then $\angle DIB = 60^\circ = \angle D_1IB$, because $\triangle DBD_1$ is isosceles. Hence, $\angle EIA = 60^\circ = \angle AIB - \angle D_1IB$. Then $\triangle AIE = \triangle AID_1$, so $EA = AD_1$ and $AE + BD = AB$.

Conversely, if $AE + BD = AB$, then $EA = AD_1$, so $\triangle AIE = \triangle AID_1$. Then $\angle EIA = \angle AID_1$. Since $\triangle DBD_1$ is isosceles we have $\angle DIB = \angle D_1IB$. Then $\angle EIA + \angle DIB = \angle AIB = \angle EID$, so $120^\circ = \angle AIB = 90^\circ + \frac{1}{2}\angle ACB \Rightarrow \angle ACB = 60^\circ$.