10.097. The area of a right-angled triangle (Fig. 10.93) is $2 \sqrt{3}$ cm $^{2}$. Determine its height drawn to the hypotenuse, if it divides the right angle in the ratio $1: 2$.
Solution.

$\angle A B D=30^{\circ}, \angle D B C=60^{\circ}$. The area of $\triangle A B C$ is found using the formula $S=\frac{1}{2} A B \cdot B C$. From $\triangle B D C$ we get $B C=\frac{D C}{\sin 60^{\circ}}$. From $\triangle A D B$ we find $A B=\frac{A D}{\sin 30^{\circ}}$. Then $S=\frac{1}{2} \frac{D C \cdot A D}{\sin 60^{\circ} \cdot \sin 30^{\circ}}=\frac{2}{\sqrt{3}} D C \cdot A D$. Since $\triangle A D B$ and $\triangle B D C$ are similar, $\frac{B D}{A D}=\frac{D C}{B D}$. Hence $B D^{2}=A D \cdot D C$. Then $S=\frac{2}{\sqrt{3}} B D^{2}=2 \sqrt{3}$ and $B D=\sqrt{3}(\mathrm{~cm})$.

Answer: $\sqrt{3}$ cm.