3.15. The height of an isosceles trapezoid is $h$, and the angle between its diagonals, opposite the lateral side, is $\alpha$. Find the midline of the trapezoid.
3.15. By the condition, in the isosceles trapezoid $A B C D$ we have: $A B=C D$, $B B_{1} \perp A D, \quad B B_{1}=h$, $A C \cap B D=O, \angle C O D=\alpha$ (Fig. 3.17). Since $\angle C O D$ is the external angle of the isosceles triangle $A O D$,

then $\angle O A D=\angle O D A=\frac{\alpha}{2}$. Next, we find $B_{1} D=E D+B_{1} E=\frac{1}{2} A D+$ $+\frac{1}{2} B C=\frac{1}{2}(A D+B C)=M N$, where $M N-$ is the midline of the trapezoid. Finally, from $\triangle B B_{1} D$ we get $B_{1} D=B B_{1} \operatorname{ctg} \frac{\alpha}{2}=h \operatorname{ctg} \frac{\alpha}{2}$.