4. For different points $A, B, C, D$ and $E$ in the plane, the following is known:

- point $A$ is equidistant from points $B, C, D$ and $E$, and these distances are 1,
- point $C$ is equidistant from points $B$ and $D$, and these distances are $\frac{1}{2}$,
- the distance between points $B$ and $E$ is 2.

Determine the remaining distances between these points.
4. From the conditions of the problem, it follows that points $B, C, D$ and $E$ belong to the same circle with center at $A$ and radius 1, and that $\overline{B E}$ is the diameter of this circle.

It is necessary to calculate $|B D|, |C E|$, and $|D E|$.

By Thales' theorem, the angle $\varangle E C B$ is a right angle, so by applying the Pythagorean theorem to $\triangle C E B$ we get:

$|C E|^{2}=|B E|^{2}-|B C|^{2}$

$|C E|^{2}=2^{2}-\left(\frac{1}{2}\right)^{2}$

$|C E|^{2}=\frac{15}{4}$

thus, $|C E|=\frac{\sqrt{15}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_90cd47adf89054fd6201g-4.jpg?height=417&width=483&top_left_y=1482&top_left_x=1255)

Furthermore, by the S-S-S congruence theorem, $\triangle A B C \cong \triangle A C D$, so $|\angle A C B|=|\angle D C A|$.

Let point $P$ be the intersection of segments $\overline{A C}$ and $\overline{B D}$. By the S-A-S congruence theorem, $\triangle B C P \cong \triangle C D P$.

This means that $|\angle B P C|=|\angle C P D|=90^{\circ}$. Therefore, $\overline{B P}$ is the altitude of triangle $\triangle A B C$.

Let $N$ be the foot of the altitude of triangle $\triangle A B C$ drawn from vertex $A$. Since $\triangle A B C$ is isosceles, we have

$|A N|^{2}=|A C|^{2}-|C N|^{2}$

$|A N|^{2}=1^{2}-\left(\frac{1}{4}\right)^{2}$

$|A N|^{2}=\frac{15}{16}$

$|A N|=\frac{\sqrt{15}}{4}$.

By the A-A similarity theorem, $\triangle A N C \sim \triangle B C P$, so

$\frac{|B P|}{|B C|}=\frac{|A N|}{|A C|}$

$|B P|=\frac{|A N|}{|A C|} \cdot|B C|$

$|B P|=\frac{1}{|A C|} \cdot|A N| \cdot|B C|$

$|B P|=\frac{1}{1} \cdot \frac{\sqrt{15}}{4} \cdot \frac{1}{2}=\frac{\sqrt{15}}{8}$.

Furthermore, $|B D|=2 \cdot|B P|=2 \cdot \frac{\sqrt{15}}{8}=\frac{\sqrt{15}}{4}$.

Finally, by Thales' theorem, the angle $\angle E D B$ is a right angle, so by applying the Pythagorean theorem to $\triangle D E B$ we get:

$|D E|^{2}=|B E|^{2}-|B D|^{2}$

$|D E|^{2}=2^{2}-\left(\frac{\sqrt{15}}{4}\right)^{2}$

$|D E|=\frac{7}{4}$.

Note: $|B P|$ could also be calculated by expressing the area of triangle $\triangle A B C$ in two ways:

$\frac{|B C| \cdot|A N|}{2}=\frac{|A C| \cdot|B P|}{2}$.