In the quadrilateral $ABCD$, measuring the geometric mean of the sides starting from $A$ on the rays $AB$ and $AD$, we get the points $K$ and $L$, and measuring the geometric mean of the sides starting from $C$ on the rays $CB$ and $CD$, we get the points $M$ and $N$. Under what condition does $KL = MN$ hold?
Let the sides of the quadrilateral $ABCD$ be denoted as $a, b, c, d$ as shown in the diagrams, and the angles at vertices $A$ and $C$ as $\alpha$ and $\gamma$, respectively.

The condition $KL = MN$ is satisfied if and only if $KL^2 - MN^2 = 0$, which, by applying the cosine rule to triangles $AKL$ and $CNM$, means

$$
\left[(\sqrt{ad})^2 + (\sqrt{ad})^2 - 2 \sqrt{ad} \cdot \sqrt{ad} \cos \alpha\right] - \left[(\sqrt{bc})^2 + (\sqrt{bc})^2 - 2 \sqrt{bc} \cdot \sqrt{bc} \cos \gamma\right] = 0
$$

However, applying the cosine rule to the common side $BD$ of triangles $ABD$ and $CBD$, we get

$$
BD^2 = a^2 + d^2 - 2ad \cos \alpha = b^2 + c^2 - 2bc \cos \gamma
$$

Therefore, condition (1) can be rewritten as follows:

$0 = (2ad - 2ad \cos \alpha) - (2bc - 2bc \cos \gamma) = 2ad - 2bc + b^2 + c^2 - (a^2 + d^2) = (b - c)^2 - (a - d)^2 = (a + b - c - d)(b + d - a - c)$

Since we have performed equivalent transformations, we can state that $KL = MN$ is satisfied if and only if $a + b = c + d$ or $a + c = b + d$. (This condition can also be stated as the points $A$ and $C$ lying on the same hyperbola with foci at $B$ and $D$.)

Based on the work of Tamás Terpai (Fazekas M. Főv. Gyak. Gimn., 12th grade)
![](https://cdn.mathpix.com/cropped/2024_05_02_f2d7f9b098a5e5c40545g-1.jpg?height=776&width=674&top_left_y=934&top_left_x=714)