5. (3 points) In triangle $A B C$, a square $K L M N$ with side length 1 is inscribed: points $K$ and $L$ lie on side $A C$, points $M$ and $N$ lie on sides $A B$ and $B C$ respectively. The area of the square is half the area of the triangle. Find the length of the height $B H$ of triangle $A B C$.
Answer: 2.

Solution 1:

$A C \cdot B H=2 S_{A B C}=4 S_{K L M N}=4$.

Points $A, L, K, C$ lie on a straight line in that exact order. Moreover, triangle $A B C$ is clearly acute. From the condition, it follows that

$$
\begin{gathered}
S_{A M L}+S_{C K N}+S_{B M N}=S_{K L M N} \\
\frac{A L \cdot 1}{2}+\frac{C K \cdot 1}{2}+\frac{(B H-1) \cdot 1}{2}=1 \\
A L+C K+B H-1=2 \\
A C-1+B H-1=2 \\
A C+B H=4=A C \cdot B H=2+\frac{A C \cdot B H}{2} . \\
2 A C+2 B H=A C+B H+4 \\
(A C-2)(B H-2)=0 \\
{\left[\begin{array}{l}
A C=2 \\
B H=2
\end{array}\right.}
\end{gathered}
$$

If only one of the equalities in this system is satisfied, then $A C \cdot B H$ will be either greater or less than 4, which is impossible. Therefore, $A C=B H=2$.

## Solution 2:

Points $A, L, K, C$ lie on a straight line in that exact order. Moreover, triangle $A B C$ is clearly acute.

On ray $L K$, take a point $X$ such that $A L=L X$. Then triangles $A M L$ and $X M L$ are equal. Similarly, on ray $K L$, take a point $Y$ such that $C K=K X$. Then triangles $C K N$ and $Y N L$ are equal. Let $Z$ be the intersection of $M X$ and $N Y$. Triangle $M N Z$ is equal to triangle $M N B$ by side $M N$ and the angles adjacent to it.

We obtain the equality:

$$
S_{K L M N}=S_{A M L}+S_{C K N}+S_{B M N}=S_{X M L}+S_{Y K N}+S_{M N Z}=S_{K L M N} \pm S_{X Y Z}
$$

where the sign $\pm$ depends on whether point $Z$ lies inside or outside triangle $A B C$. This area must be equal to 0, i.e., point $Z$ lies on $A C$. From this, the height of triangle $Z M N$, dropped from $Z$ to $M N$, is equal to the side of the square. But this height is equal to the height of triangle $B M N$, dropped from $B$ to $M N$. If we add the side of the square to this height, we get the height $B H$, from which $B H=2$.