4. Analysis. Let $A^{\prime}$ and $B^{\prime}$ be the feet of the perpendiculars from $A$ and $B$, respectively, to the opposite sides, $A_{1}$ the midpoint of $B C$, and let $D^{\prime}$ be the foot of the perpendicular from $A_{1}$ to $A C$. We then have $A A_{1}=m_{a}$, $A A^{\prime}=h_{a}, \angle A A^{\prime} A_{1}=90^{\circ}, A_{1} D^{\prime}=h_{b} / 2$, and $\angle A D^{\prime} A_{1}=90^{\circ}$. Construction. We construct the quadrilateral $A D^{\prime} A_{1} A^{\prime}$ (starting from the circle with diameter $A A_{1}$ ). Then $C$ is the intersection of $A^{\prime} A_{1}$ and $A D^{\prime}$, and $B$ is on the line $A_{1} C$ such that $C A_{1}=A_{1} B$ and $\mathcal{B}\left(B, A_{1}, C\right)$. Discussion. We must have $m_{a} \geq h_{a}$ and $m_{a} \geq h_{b} / 2$. The number of solutions is 0 if $m_{a}=h_{a}=h_{b} / 2,1$ if two of $m_{a}, h_{a}, h_{b} / 2$ are equal, and 2 otherwise.