[ Theorem of Three Perpendiculars $\left[\begin{array}{l}\text { O } \\ {[\text { Orthocentric tetrahedron }}\end{array}\right]$

The height of the pyramid $A B C D$, dropped from the vertex $D$, passes through the point of intersection of the altitudes of the triangle $A B C$. In addition, it is known that $D B=b, D C=c, \angle B D C=90^{\circ}$. Find the ratio of the areas of the faces $A D B$ and $A D C$.

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Let $M$ be the point of intersection of the altitudes of triangle $ABC$, $DM$ be the height of the pyramid, and the lines $BM$ and $AC$ intersect at point $K$. Then $BK$ is the height of triangle $ABC$, and since line $BK$ is the orthogonal projection of the slant $DB$ on the base plane $ABC$, by the theorem of three perpendiculars, $DB \perp AC$. Therefore, line $DB$ is perpendicular to two intersecting lines $AC$ and $DC$ in the plane of face $ACD$. Hence, $\angle ADB = 90^\circ$. Similarly, $\angle ADC = 90^\circ$. Therefore,

$$
\frac{S_{\triangle ADB}}{S_{\triangle ADC}} = \frac{\frac{1}{2} AD \cdot DB}{\frac{1}{2} AD \cdot DC} = \frac{DB}{DC} = \frac{b}{a}
$$

## Answer

$\frac{b}{c}$

All faces of the pyramid $PQRS$ are acute triangles, and the perpendiculars dropped from vertices $P, Q, R, S$ to the opposite faces are equal. It is known that $SP = 6$, $\angle SRQ = 75^\circ$, and $\angle SPR = 45^\circ$. Find the edge $PQ$.

## Answer

$2 \sqrt{6}$.