In the diagram at right, $A B C D$ is a parallelogram. Given that the distances from $A, B$, and $C$ to line $\ell$ are, respectively, 3,1 , and 5 , find the distance from $D$ to $\ell$ and prove that your answer is correct.

Remark. In the contest, the labels on the points $A$ and $C$ were switched. This makes the figure grossly out of scale but does not change the solution.
As shown, label the feet of the perpendiculars from $A, B, C, D$ to $\ell$ by $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$. Also, drop perpendiculars $B X$ and $C Y$ to $A A^{\prime}$ and $D D^{\prime}$ respectively. We find that triangles $A B X$ and $D C Y$ are similar since their corresponding sides are parallel, and in fact congruent since $A B$ and $C D$ are opposite sides of the parallelogram. Thus $A X=D Y$. But $A X=A A^{\prime}-$ $X A^{\prime}=A A^{\prime}-B B^{\prime}=3-1=2$, and hence $D D^{\prime}=D Y+Y D^{\prime}=A B+C C^{\prime}=$

![](https://cdn.mathpix.com/cropped/2024_04_17_89d84300f5630edac230g-1.jpg?height=472&width=572&top_left_y=534&top_left_x=1335)
$2+5=7$.