12.005. 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.
## Solution.

Given that in the isosceles trapezoid $ABCD$ we have: $AB = CD$, $BB_1 \perp AD, BB_1 = h, AC \cap BD = O, \angle COD = \alpha$ (Fig. 12.8). Since $\angle COD$ is the external angle of the isosceles triangle $AOD$, we have $\angle OAD = \angle ODA = \frac{\alpha}{2}$. Further, we have $B_1D = ED + B_1E = \frac{1}{2} AD + \frac{1}{2} BC = \frac{1}{2}(AD + BC) = MN$, where $MN$ is the midline of the trapezoid. From $\triangle BB_1D$ we get $B_1D = BB_1 \operatorname{ctg} \frac{\alpha}{2} = h \operatorname{ctg} \frac{\alpha}{2}$.

Answer: $h \operatorname{ctg} \frac{\alpha}{2}$.