The vertices of a quadrilateral are $A(0 ; 0), B(0 ;-1), C(1 ; 0), D\left(\frac{\sqrt{2}}{2} ; \frac{\sqrt{2}}{2}\right)$. Let $K$ be the intersection point of the lines $AB$ and $CD$. Calculate the distance $AK$.
Notice that $BC$ is parallel to $AD$ (both segments have a slope of 1), so triangles $BCK$ and $ADK$ are similar.

![](https://cdn.mathpix.com/cropped/2024_05_02_40b404deffcee7078438g-1.jpg?height=626&width=290&top_left_y=259&top_left_x=904)

The ratio of the corresponding sides is equal, thus

$$
\frac{BK}{AK}=\frac{BC}{AD}
$$

Since $BK=AB+AK$, and $AB=1, \quad AD=1, \quad BC=\sqrt{2}$, we have

$$
AK=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1
$$

Krisztina Jambrich (Szekszárd, Garay J. Gymnasium, 4th grade)