10.229. A parallelogram with sides of 3 cm and 5 cm and a diagonal of 6 cm is inscribed in a triangle. Find the sides of the triangle, given that the diagonals of the parallelogram are parallel to the lateral sides of the triangle, and the smaller of its sides lies on the base of the triangle.
Solution.

Let in $\triangle A B C D K=E F=3 \text{cm}, D E=5 \text{cm}, D F=6$ cm, $D K \| E F$, $O$ - the point of intersection of the diagonals $D F$ and $E K$ of the parallelogram $D E F K$ (Fig. 10.38). Let $E O=x$ cm. Then $D F^{2}+E K^{2}=2\left(D E^{2}+E F^{2}\right)$; $36+4 x^{2}=2(25+9) ; x=2 \sqrt{2}$. OEBF is a parallelogram. Then $B E=F O=3 \text{cm}, B F=E O=2 \sqrt{2} \text{cm}$. K E F C is a parallelogram. Then $F C=E K=4 \sqrt{2}$ cm, $K C=E F=3 \text{cm}$. A E F D is a parallelogram. Then $A E=D F=6$ cm, $A D=E F=3$ cm. Therefore, $A B=A E+E B=9$ cm, $B C=B F+F C=6 \sqrt{2} \text{cm}, A C=A D+D K+K C=9 \text{cm}$.

Answer: 9 cm; 9 cm; $6 \sqrt{2}$ cm.