10.355. An isosceles triangle with sides 8.5 and 5 is divided into three equal areas by perpendiculars drawn from a certain point to its sides.

Find the distance from this point to each side of the given triangle.
## Solution.

Let $B K$ be the height of $\triangle A B C, A B=B C$ (Fig. 10.136.1). Points $D, E$, $F_{1}$ are the bases of the perpendiculars dropped from point $F$ to the sides of the triangle, dividing it into three equal areas. We will prove, using proof by contradiction, that point $F$ lies on $B K$, and point $F_{1}$ coincides with point $K$. Suppose $A F_{1}>F_{1} C$. Draw a segment $M N$ through point $F$ parallel to $A C$. Then $M F>F N$. $S_{A F_{1} F D}=S_{A M F F_{1}^{1}}+S_{\triangle M D F}, S_{C F_{1} F E}=S_{C N F F_{1}}+S_{\triangle N E F}$. Since $M F>F N, A F_{1}>F_{1} C$, and $F F_{1}$ is the common height of trapezoids $A M F F_{1}$ and $C N F F_{1}$, then $S_{A M F F_{1}}>S_{C N F F_{1}}$. Since $\triangle M D F \sim \triangle N E F\left(\angle M D F=\angle N E F=90^{\circ}, \angle D M F=\angle E N F=\angle A\right)$ and $M F>F N$, then $S_{\triangle M D F}>S_{\triangle N E F}$. Therefore, $S_{A F_{1} F D}>S_{C F_{1} F E}$, which contradicts the condition. Thus, point $F$ lies on $B K$ (Fig. 10.136.2). From
$\triangle A K B\left(\angle A K B=90^{\circ}\right): B K=\sqrt{A B^{2}-A K^{2}}=3 . S_{\triangle A B C}=\frac{1}{2} A C \cdot B K=12$.

$S_{F D B E}=\frac{1}{3} S_{\triangle A B C}=4 \cdot S_{\triangle F D B}=\frac{1}{2} S_{F D B E}=2 \cdot$ Let $F D=x, D B=y$.

$F D \cdot D B=2 S_{\triangle F D B}=4, x y=4$.

From $\angle A B K: \frac{F D}{D B}=\frac{A K}{B K}=\frac{4}{3}, \frac{x}{y}=\frac{4}{3}$.

We get the system of equations

$$
\left\{\begin{array}{l}
x y=4, \\
\frac{x}{y}=\frac{4}{3} .
\end{array} \text { Then } x=\frac{4}{3} y, \frac{4}{3} y^{2}=4, y=\sqrt{3}, x=\frac{4 \sqrt{3}}{3}\right. \text {. }
$$

From $\triangle B D F\left(\angle B D F=90^{\circ}\right): B F=\sqrt{F D^{2}+D B^{2}}=\sqrt{\frac{16}{3}+3}=\frac{5 \sqrt{3}}{3}$. $F K=B K-B F=3-\frac{5 \sqrt{3}}{3}=\frac{9-5 \sqrt{3}}{3}$.

Answer: $\frac{4 \sqrt{3}}{3} ; \frac{4 \sqrt{3}}{3} ; \frac{9-5 \sqrt{3}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0813.jpg?height=492&width=666&top_left_y=1155&top_left_x=351)

Fig. 10.137