8,9

In a parallelogram, the ratio of the sides and the ratio of the diagonals are the same and equal to $\frac{1}{2}$. From the vertex of the obtuse angle A, a height $A E$ is dropped to the larger side $C D$. Find the ratio $\frac{D E}{C E}$.
Let $ABCD$ be a parallelogram with an obtuse angle at vertex $A$. Let $AD=a, CD=2a, AC=d$. Since the larger diagonal of a parallelogram lies opposite the obtuse angle, $BD=2d$. The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of all its sides, so

$$
2a^2 + 2 \cdot 4a^2 = d^2 + 4d^2
$$

from which $d^2 = 2a^2$. Let $\angle ADC = \alpha$. From triangle $ADC$, by the cosine rule, we find that

$$
\cos \alpha = \frac{AD^2 + DC^2 - AC^2}{2AD \cdot DC} = \frac{a^2 + 4a^2 - 2a^2}{4a^2} = \frac{3}{4}
$$

Then

$$
DE = AD \cos \alpha = \frac{3}{4}a, EC = DC - DE = 2a - \frac{3}{4}a = \frac{5}{4}a.
$$

Therefore, $\frac{DE}{CE} = \frac{3}{CE} = \frac{3}{1} = \frac{3}{5}$.

## Answer

$\frac{3}{5}$.