232. Inscribe a square in an equilateral triangle so that one of its sides lies on the base of the triangle.
232. 233) Let the square $D E F H$ be the one sought (Fig. 54). Let its side be $D E = x$. We have $D E: B K = A C: B O$; if $A C = a$, then $B O = \frac{a}{2} \sqrt{3}$, hence

$$
x: \left(\frac{a \sqrt{3}}{2} - x\right) = a: \frac{a \sqrt{3}}{2} = 1: \frac{\sqrt{3}}{2},
$$

from which

$$
x = a(2 \sqrt{\overline{3}} - 3)
$$

2) Geometric construction solution. Connect point $E$ with $A$, assuming the sought square is constructed. Through an arbitrary point $e$ on line $A E$, draw $d e \| D E$ and $e f \| E F$. Since $d e: D E = A e: A E = e f: E F$, it follows from the equality of $D E$ and $E F$

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-132.jpg?height=460&width=502&top_left_y=1780&top_left_x=366)

Fig. 54. that $d e = e t$, i.e., figure $e f h d$ is a square. Hence the construction: taking an arbitrary square $d e f h$, connect $A$ with point $e$ and extend $A e$ until it meets $B C$ at point $E$. Through $E$, draw $E D \| d e. ~ D E$ is the side of the sought square.