10.208. Perpendiculars are drawn from the vertex of the acute angle of a rhombus to the lines containing the sides of the rhombus to which this vertex does not belong. The length of each perpendicular is 3 cm, and the distance between their bases is $3 \sqrt{3}$ cm. Calculate the lengths of the diagonals of the rhombus.
Solution.

Since $\triangle A E F$ is isosceles (Fig. 10.18), the bisector $A M$ is perpendicular to $E F$ and lies on the diagonal of the rhombus. We find $A M^{2}=A F^{2}-M F^{2}=9-\frac{27}{4}=\frac{9}{4}$, i.e., $A M=\frac{3}{2}$ (cm). In $\triangle A C F$, we have $\angle F=90^{\circ}$ and $F M \perp A C \Rightarrow A F^{2}=A C \cdot A M, 9=A C \cdot \frac{3}{2}, A C=6$ (cm).

Next, $\triangle A C D \sim \triangle A E F$ (the base angles are equal as angles with mutually perpendicular sides), so $\frac{A M}{O D}=\frac{E F}{A C} \Leftrightarrow \frac{3 / 2}{O D}=\frac{3 \sqrt{3}}{6}$, $O D=\sqrt{3}$ cm. We get $B D=2 \sqrt{3}$ cm, $A C=6$ cm.

Answer: 6 and $2 \sqrt{3}$ cm.