Exercise 14. Let $A B C D$ be a rectangle and $M$ the midpoint of the segment $[C D]$. A line parallel to $(A B)$ intersects the segments $[A D],[A M],[B M],[B C]$ at points $P, Q, R$ and $S$, respectively. The line $(D R)$ intersects the segment $[A M]$ at $X$ and the segment $[B C]$ at $Y$. If $D X=6$ and $X R=4$, what is the length of the segment $[R Y]$?
## Solution alternative $n^{\circ} 1$

![](https://cdn.mathpix.com/cropped/2024_06_04_17def46ea4e6119381e4g-26.jpg?height=1263&width=1311&top_left_y=128&top_left_x=378)

We propose a solution using Thales' theorem in two other "butterflies."

We introduce $E$ as the intersection point of the lines $(A M)$ and $(B C)$, and $F$ as the intersection point of the lines $(B M)$ and $(A D)$. Since $M C = \frac{1}{2} A B$, the line $(M C)$ is a midline for the triangle $A B E$. Thus, $B C = C E$ and $A M = M E$. Similarly, $A D = D F$ and $B M = M F$.

We now set $x = R Y$ and will seek two equations involving $X$. On the one hand, according to Thales' theorem in the butterfly $B Y R D F$, we have

$$
\frac{x}{10} = \frac{R Y}{R D} = \frac{B Y}{D F} = \frac{B Y}{A D}
$$

On the other hand, according to Thales' theorem in the butterfly $A D X Y E$, we find

$$
\frac{4 + x}{6} = \frac{X Y}{D X} = \frac{Y E}{A D} = \frac{B E - B Y}{A D} = 2 - \frac{B Y}{A D}
$$

By combining these two equations, we get

$$
\frac{2}{3} + \frac{x}{6} = 2 - \frac{x}{10}
$$

which has the solution $x = 5$.