12.138. The larger base of a trapezoid inscribed in a circle is equal to the diameter of the circle, and the angle at the base is $\alpha$. In what ratio does the point of intersection of the diagonals of the trapezoid divide its height?
Solution.

Let the base $AD$ of the isosceles trapezoid $ABCD$ (Fig. 12.11) be the diameter of the circle circumscribed around the trapezoid, then the center $O$ of the circle is the midpoint of $AD$.

The height $KO$ of the trapezoid passes through the point $L$ of intersection of the diagonals, $\triangle BLC \sim \triangle ALD, \frac{KL}{LO}=\frac{LC}{LD}$.

$\angle ACD$ is an inscribed angle subtending the diameter, so $\angle ACD=90^{\circ}$.

$\angle ACD$ and $\angle AOL$ are right angles with a common acute angle at vertex $A$. Therefore, $\angle ALO=\angle ADC=\alpha$.

Then $\angle KLC=\angle OLD=\alpha, \angle CLD=180^{\circ}-2\alpha$, from $\triangle DLC \left(\angle LCD=90^{\circ}\right)$:

$$
\frac{LC}{LD}=\cos \angle CLD=\cos \left(180^{\circ}-2\alpha\right)=-\cos 2\alpha
$$

Answer: $-\cos 2\alpha$.