B2. For a quadrilateral $A B C D$, the angles at $A$ and $B$ are right angles. Furthermore, $|A B|=5$ and $|A D|=|C D|=6$.

Determine all possible values of $|B C|$.
B2. $6 \pm \sqrt{11}$

Consider line segment $AB$ with length 5 and two lines perpendicular to it through points $A$ and $B$. Point $D$ lies on the line through $A$ at a distance of 6 from $A$, and point $C$ lies on the line through $B$ on the same side of $AB$ as point $D$. Since $|CD|=6$, point $C$ lies on the circle centered at $D$ with radius 6. This gives two possible positions for point $C$, which we denote as $C_{1}$ and $C_{2}$, as shown in the figure. Now take the perpendicular projections of $C_{1}$ and $C_{2}$ onto $AD$ and call them $E_{1}$ and $E_{2}$, respectively. The Pythagorean theorem tells us that

$$
\left|D E_{1}\right|^{2}=\left|D C_{1}\right|^{2}-\left|C_{1} E_{1}\right|^{2}=6^{2}-5^{2}=11
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_cb88577c1f7dc72a0c75g-1.jpg?height=440&width=303&top_left_y=1422&top_left_x=1505)

Thus, $\left|D E_{1}\right|=\sqrt{11}$ and $\left|B C_{1}\right|=\left|A E_{1}\right|=|A D|-\left|D E_{1}\right|=6-\sqrt{11}$. In exactly the same way, we find that $\left|D E_{2}\right|=\sqrt{11}$ and thus $\left|B C_{2}\right|=\left|A E_{2}\right|=|A D|+\left|D E_{2}\right|=6+\sqrt{11}$.

The possible values for $|BC|$ are therefore $6-\sqrt{11}$ and $6+\sqrt{11}$.