Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_22fa77b82e311267ee30g-26.jpg?height=327&width=397&top_left_y=95&top_left_x=526)
Answer: 7.

Solution. Since $A B C D$ is a square, then $A B=B C=C D=A D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_22fa77b82e311267ee30g-26.jpg?height=333&width=397&top_left_y=584&top_left_x=526)

Fig. 1: to the solution of problem 8.4

Notice that $\angle A B K=\angle C B L$, since they both complement $\angle A B L$ to $90^{\circ}$. Then the right triangles $A B K$ and $C B L$ are equal by the acute angle and the leg $A B=B C$ (Fig. 1). Therefore, $A K=C L=6$. Then

$$
L D=C D-C L=A D-C L=(K D-A K)-C L=K D-2 \cdot C L=19-2 \cdot 6=7
$$