Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5b323dccc142e96a9907g-30.jpg?height=426&width=515&top_left_y=1495&top_left_x=469)
Answer: 13.

![](https://cdn.mathpix.com/cropped/2024_05_06_5b323dccc142e96a9907g-31.jpg?height=431&width=519&top_left_y=166&top_left_x=467)

Fig. 5: to the solution of problem 9.4

Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute angle $\angle Y A D=\angle X B A$. The similarity ratio is the ratio $A B: A D$, which is $\frac{1}{2}$. From this, we get $X A=\frac{1}{2} D Y=5$ and $A Y=2 B X=8$, which in total gives $X Y=13$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5b323dccc142e96a9907g-31.jpg?height=428&width=517&top_left_y=947&top_left_x=468)

Fig. 6: to the solution of problem 9.4

Another solution. Let $M$ be the midpoint of segment $A D$, then $A M=M D=\frac{A D}{2}=\frac{B C}{2}=A B$ (Fig. 6). Drop a perpendicular from $M$ to $M H$ on line $\ell$. Since in triangle $A D Y$ the segment $M H$ passes through the midpoint of side $A D$ and is parallel to side $D Y$, it is the midline, $A H=H Y$ and $M H=\frac{D Y}{2}=\frac{10}{2}=5$.

Note that $\angle A B X=90^{\circ}-\angle B A X=90^{\circ}-\left(180-90^{\circ}-\angle M A H\right)=\angle M A H$. Then right triangles $A B X$ and $M A H$ are equal by hypotenuse $A B=A M$ and acute angle $\angle A B X=\angle M A H$. From this, it follows that $A X=M H=5$ and $A H=B X=4$.

Thus, $X Y=A X+A H+H Y=5+4+4=13$.