Let $A B C D$ be a rectangle such that $A B=\sqrt{2} B C$. Let $E$ be a point on the semicircle with diameter $A B$, as indicated in the following figure. Let $K$ and $L$ be the intersections of $A B$ with $E D$ and $E C$, respectively. If $A K=2 \mathrm{~cm}$ and $B L=9 \mathrm{~cm}$, calculate, in $\mathrm{cm}$, the length of the segment $K L$.

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-30.jpg?height=412&width=495&top_left_y=669&top_left_x=667)
Solution

Let $x$ and $y$ be the lengths of the orthogonal projections of segments $E K$ and $E L$ onto segment $A B$, and $P$ be the orthogonal projection of $E$ onto $A B$. Also, let $h = E P$. By the similarity of triangles, we have

$$
\frac{2}{x} = \frac{B C}{h} \text{ and } \frac{9}{y} = \frac{B C}{h}
$$

Therefore,

$$
\begin{aligned}
K L & = x + y \\
& = \frac{11 h}{B C}
\end{aligned}
$$

Since $E A B$ is a right triangle at $E$, it follows from the metric relations of right triangles that

$$
\begin{aligned}
h^{2} & = A P \cdot P B \\
& = (2 + x)(9 + y) \\
& = \left(2 + \frac{2 h}{B C}\right)\left(9 + \frac{9 h}{B C}\right) \\
& = 18\left(1 + \frac{h}{B C}\right)^{2}
\end{aligned}
$$

Therefore,

$$
1 + \frac{h}{B C} = \frac{h}{3 \sqrt{2}}
$$

Moreover, from

it follows that

$$
\sqrt{2} B C = A K + K L + L B = 11\left(1 + \frac{h}{B C}\right)
$$

$$
\frac{h}{3 \sqrt{2}} = \frac{\sqrt{2} B C}{11}
$$

That is, $\frac{h}{B C} = \frac{6}{11}$ and $K L = \frac{11 h}{B C} = 6 \text{ cm}$.