70. The base of the pyramid $A B C D E$ is a parallelogram $A B C D$. No lateral face is an obtuse triangle. There exists a point $M$ on the edge $D C$ such that the line $E M$ is perpendicular to $B C$. In addition, the diagonal of the base $A C$ and the lateral edges $E D$ and $E B$ are related by the following ratios: $|A C| \geqslant \frac{5}{4}|E B| \geqslant \frac{5}{3}|E D|$. A section is made through vertex $B$ and the midpoint of one of the lateral edges, forming an isosceles trapezoid. Find the ratio of the area of the section to the area of the base of the pyramid.
70. First, we will prove that $A B C D$ is a rectangle and the plane $D E C$ is perpendicular to the plane $A B C D$. For this, we will draw a section through $E$ perpendicular to $B C$. This section must intersect the base along a line passing through $M$ and intersecting the segments $B C$ and $A D$ (possibly at their endpoints). Next, a section through $B$ that is an isosceles trapezoid can only be drawn if this section contains the edge $A B$, and $|D E|=|E C|, |A E|=|E B|$. Therefore,

$$
\frac{3}{5}|A C| \geqslant|E D|=|E C|, \quad \frac{4}{5}|A C| \geqslant|E B|=|A E|,
$$

i.e., $|A C|^{2} \geqslant|C E|^{2}+|A E|^{2}$ and $\triangle A E C$ is not an acute triangle. However, $\widehat{A E C}$ cannot be obtuse, otherwise $\widehat{D E C}$ would be obtuse.

Thus, $|A C|=\frac{5}{4}|A E|=\frac{5}{3}|E C|$.

Answer: $\frac{3}{\delta} \sqrt{\frac{65}{14}}$.