110. All edges of the triangular pyramid $A B C D$ touch a certain sphere. Three segments connecting the midpoints of the skew edges are equal. The angle $A B C$ is $100^{\circ}$. Find the ratio of the heights of the pyramid dropped from vertices $A$ and $B$.
110. From the fact that the edges of the pyramid $ABCD$ touch the sphere, it follows that the sums of opposite edges of the pyramid are equal. Extend the pyramid $ABCD$ to a parallelepiped by drawing a plane through each edge of the pyramid, parallel to the opposite edge. The edges of the pyramid will be the diagonals of the faces of the parallelepiped (Fig. 17), and the edges of the parallelepiped will be equal to the distances between the midpoints of opposite edges of the pyramid. Let $|AD|=a$, $|BC|=b$, then any two opposite edges of the pyramid will be equal to $a$ and $b$. We will prove this. Let $|AB|=x,|DC|=y$. Then $x+y=a+b, x^{2}+y^{2}=$ $=a^{3}+b^{2}$ (the last equality follows from the fact that all faces of the parallelepiped are rhombuses with equal sides).

From this, it follows that $x=a, y=b$ or $x=b, y=a$. Therefore, in $\triangle ABC$ at least two sides are equal to each other. But $\widehat{ABC}=100^{\circ}$, so $|AB|=x=|BC|=b,|AC|=a$, $|DB|=b,|DC|=a$.

From $\triangle ABC$ we find $a=2 b \sin 50^{\circ}$,

$$
\begin{aligned}
& V_{ABCD}=\frac{1}{3} S_{ADC^{h_{B}}}=\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4} h_{B}= \\
& =\frac{1}{3} S_{DBC^{h} A}=\frac{1}{3} \cdot \frac{b^{2} \sin 100^{\circ}}{2} h_{A},
\end{aligned}
$$

from which $\frac{h_{A}}{h_{B}}=\frac{a^{2} \sqrt{3}}{2 b^{2} \sin 100^{\circ}}=\sqrt{\overline{3}} \operatorname{tg} 50^{\circ}$.