Construct triangle $ABC$ if one side $BC$, one altitude $AA_1$, and the fact that the angle bisector of $\angle A$ is the median of triangle $AA_1D$, where $D$ is the midpoint of side $BC$, are given.
Let $AE$ be the median of the triangle $AA_1D$. If we draw a perpendicular to $BC$ at $D$, it intersects $AE$ at point $F$, which lies on the circumference of the circumcircle of the triangle, because $AE$ bisects angle $A$ and arc $BC$. Since $DE = EA_1$, $\angle DEF = \angle AEA_1$, and $\angle FDE = \angle EA_1A = 90^\circ$, therefore $\triangle DFE \cong \triangle A_1AE$ and thus $DF = AA_1$. Based on this, the construction is as follows:

On the given side $BC$, at its midpoint $D$, we draw a perpendicular and measure the given height $AA_1$ from point $D$ on both sides of this perpendicular. This gives us points $F$ and $G$. We draw a circle through points $B$, $C$, and $F$, which intersects the line parallel to $BC$ through point $G$ at the third vertex of the triangle. The problem is solvable if $BC \geq 2AA_1$.