5. In an acute triangle $A B C$, let $D$ be the foot of the altitude from vertex $C$ and $P, Q$ be the feet of the perpendiculars from point $D$ to sides $A C$ and $B C$. The areas of triangles $A D P, D C P, D B Q, C D Q$ are denoted by $S_{1}, S_{2}, S_{3}, S_{4}$, respectively. Calculate $S_{1}: S_{3}$, given that $S_{1}: S_{2}=2: 3, S_{3}: S_{4}=3: 8$.
SOLUTION. Let $x=|A D|, y=|B D|, v=|C D|$ (Fig. 3). From the similarity of triangles $A D P$ and $D C P$, it follows that $x^{2}: v^{2}=S_{1}: S_{2}=2: 3$. Similarly, from the similarity of triangles $D B Q, C D Q$, it follows that $y^{2}: v^{2}=S_{3}: S_{4}=3: 8$. Therefore, $x^{2}: y^{2}=(2 \cdot 8):(3 \cdot 3)=16: 9, x: y=4: 3$. Triangles $A D C, D B C$ have a common height, so $\left(S_{1}+S_{2}\right):\left(S_{3}+S_{4}\right)=x: y=4: 3$. Substituting $S_{2}=\frac{3}{2} S_{1}$ and $S_{4}=\frac{8}{3} S_{3}$, and after simplification, we get $S_{1}: S_{3}=88: 45$.