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_564c13f715a760703913g-26.jpg?height=327&width=397&top_left_y=95&top_left_x=526)
Answer: 7.

Solution. Since $ABCD$ is a square, then $AB = BC = CD = AD$.

![](https://cdn.mathpix.com/cropped/2024_05_06_564c13f715a760703913g-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 $\angle ABK = \angle CBL$, since they both complement $\angle ABL$ to $90^{\circ}$. Therefore, right triangles $ABK$ and $CBL$ are congruent by the acute angle and the leg $AB = BC$ (Fig. 1). Consequently, $AK = CL = 6$. Then

$$
LD = CD - CL = AD - CL = (KD - AK) - CL = KD - 2 \cdot CL = 19 - 2 \cdot 6 = 7
$$