4.2. Given a rectangular trapezoid $A B C D (B C \| A D$ and $A B \perp A D)$, the diagonals of which are perpendicular and intersect at point $M$. Points $K$ and $L$ are chosen on sides $A B$ and $C D$ respectively, such that $M K$ is perpendicular to $C D$, and $M L$ is perpendicular to $A B$. It is known that $A B=1$. Find the length of the segment $K L$.
4.2. Solution. Let $P$ be the intersection point of $ML$ and $AB$ (see Fig. 21). Triangles $AKM$ and $MLD$ are similar because $\angle KAM = \angle LMD$ (they complement $\angle ABM$ to a right angle) and $\angle KMA = \angle LDM$ (they complement $\angle MCD$ to a right angle),

![](https://cdn.mathpix.com/cropped/2024_06_04_b1380f2021d62732a78bg-23.jpg?height=703&width=994&top_left_y=54&top_left_x=537)

Fig. 21: to the solution of problem 4.2

thus $\frac{AK}{AM} = \frac{ML}{MD}$. Also, from the similarity of triangles $AMD$ and $BPM$, we get the equality $\frac{AM}{MD} = \frac{BP}{PM}$. These two equalities, along with the obvious equality $PM = ML$, give us the equality of segments $BP = AK$.

Let $R$ be the midpoint of segment $PK$. Then $R$ is also the midpoint of segment $AB$. $RM$ is the midline of triangle $PKL$ and the median of the right triangle $ABM$, and the length of segment $RM$ is half the length of segments $AB$ and $KL$. Therefore, $KL = AB = 1$.