12.259. The side of the lower base of a regular truncated quadrilateral pyramid is equal to $a$, the side of the upper base is equal to $b$. The lateral face forms an angle $\alpha$ with the base plane. A plane is drawn through the side of the lower base and the midpoint of the segment connecting the centers of the bases, intersecting the opposite lateral face along a certain line. Find the distance from this line to the lower base.
Solution.

Let point $P$ be the midpoint of segment $O O_{1}$, connecting the centers of the bases of the regular truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}, A D=\mathrm{a}, A_{1} D_{1}=\mathrm{b}$ (Fig. 12.125), $M N$ - the line where the plane passing through side $C D$ of the lower base and point $P$ intersects the plane of the face $A A_{1} B_{1} B$. Since $C D \| A B$, then $M N \| A B$ and line $M N$ is parallel to the plane of the base.

$E$ - the midpoint of $A B, E_{1}$ - the midpoint of $A_{1} B_{1}, F$ - the midpoint of $C D, F_{1}$ - the midpoint of $C_{1} D_{1}$. Then the plane of the quadrilateral $E E_{1} F_{1} F_{\text {p }}$ is perpendicular to the plane of the base, $\angle E E_{1} E O=\angle F_{1} F O=\alpha$.

$K$ - the point of intersection of $M N$ and $E E_{1}$. The length of the perpendicular $K L$, dropped from point $K$ to the lower base, is the desired distance.

Draw a perpendicular $F_{1} S$ from point $F_{1}$ to the lower base.

$$
S F=\frac{E F-E_{1} F_{1}}{2}=\frac{a-b}{2}
$$

In $\triangle F_{1} S F\left(\angle F_{1} S F=90^{\circ}\right): F_{1} S=S F \operatorname{tg} \angle F_{1} F S=\frac{a-b}{2} \operatorname{tg} \alpha$.

$O P=\frac{1}{2} O O_{1}=\frac{1}{2} F_{1} S=\frac{a-b}{4} \operatorname{tg} \alpha$.

If $K L=x$, then from

$\triangle K L E\left(\angle K L E=90^{\circ}\right): E L=K L \operatorname{ctg} \angle K E L=x \operatorname{ctg} \alpha$.

$L F=E F-E L=a-x \operatorname{ctg} \alpha$.

$\triangle K L F \sim \triangle P O F \Rightarrow$

$\Delta K L F \sim \Delta P O F \Rightarrow \frac{K L}{L F}=\frac{P O}{O F} ; \frac{x}{a-x \operatorname{ctg} \alpha}=\frac{\frac{a-b}{4} \operatorname{tg} \alpha}{\frac{a}{2}} \Leftrightarrow$
$\Leftrightarrow 2 a x=(a-x \operatorname{ctg} \alpha)(a-b) \operatorname{tg} \alpha \Leftrightarrow 2 a x=a^{2} \operatorname{tg} \alpha-a b \operatorname{tg} \alpha-a x+b x \Leftrightarrow$
$\Leftrightarrow x(3 a-b)=a(a-b) \operatorname{tg} \alpha ; x=\frac{a(a-b) \operatorname{tg} \alpha}{3 a-b}$.

Answer: $\frac{a(a-b) \operatorname{tg} \alpha}{3 a-b}$.