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[ Auxiliary area. The area helps to solve the problem ]
[ Thales' theorem and the theorem of proportional segments ]

In a right triangle with legs of 6 and 8, a square is inscribed, sharing a right angle with the triangle. Find the side of the square.

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Consider similar triangles or apply the method of areas.

## Solution

Let the vertex $M$ of the square $CKML$ lie on the hypotenuse $AB$ of the right triangle $ABC$, and the vertices $K$ and $L$ lie on the legs $AC$ and $BC$ respectively,

$AC=6, BC=8$.

First method. Since $CM$ is the angle bisector of angle $C$, then $AM: BM = AC: BC = 3: 4$. By Thales' theorem, $KC = \frac{4}{7} AC = \frac{24}{7}$.

Second method. Let the side of the square be $x$. Then $S_{ABC} = S_{AMC} + S_{BMC}$, $AC \cdot BC = AC \cdot MK + BC \cdot ML$, $6 \cdot 8 = 6x + 8x$, from which $x = \frac{24}{7}$.

## Answer

$\frac{24}{7}$.