[ Signs of Perpendicularity]

The height of the triangular pyramid $A B C D$, dropped from vertex $D$, passes through the point of intersection of the heights of triangle $A B C$. In addition, it is known that $D B=3, D C=2, \angle B D C=90^{\circ}$. Find the ratio of the area of face $A D B$ to the area of face $A D C$.

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Let $H$ be the point of intersection of the altitudes of triangle $ABC$. According to the problem, $DH$ is the height of the pyramid. Then $BH$ is the orthogonal projection of the slant $DB$ onto the base plane $ABC$ of the pyramid. Since $BH \perp AC$, by the theorem of three perpendiculars, $DB \perp AC$. Thus, the line $BD$ is perpendicular to two intersecting lines $DC$ and $AC$ in the plane of the face $ADC$. Therefore, the line $BD$ is perpendicular to the plane of the face $ADC$. Hence, $BD \perp AD$, i.e., triangle $ADB$ is a right triangle. Similarly, we can prove that triangle $ADC$ is also a right triangle, with $\angle ADC = 90^\circ$. Therefore,

$$
S(ADB) / S(ADC) = \left(\frac{1}{2} AD \cdot BD\right) / \left(\frac{1}{2} AD \cdot CD\right) = BD / CD = 3 / 2
$$