5. In a circle with center $O$, a diameter $M N$ is drawn, and points $K$ - the midpoint of arc $M N$, point $E$ - the midpoint of chord $M K$, and point $B$ - the midpoint of arc $K N$ are marked. Chord $A B$ is drawn through point $E$. On segment $A B$, as a side, a rectangle $A B C D$ is constructed such that its vertices $C$ and $D$ also lie on the circle. Find the area of rectangle $A B C D$ if the diameter of the circle is $d$.
Solution. Draw radii $O K$ and $O B$, then triangle $M O K$ is a right and isosceles triangle. Segment $E O$ is its median, altitude, and bisector, so the angles $\angle M O E=\angle E O K=\angle K O B=\angle B O N=45^{\circ}$. Angle $\angle B O E=\angle E O K+\angle K O B=45^{\circ}+45^{\circ}=90^{\circ}$, which means triangle $B O E$ is a right triangle. If the diameter of the circle is $d$, then $O B=\frac{d}{2}$, and $E O=\frac{d \sqrt{2}}{4}$. In the right triangle $B O E$, side $B E$ is the hypotenuse, so $B E=\sqrt{E O^{2}+B O^{2}}=\sqrt{\left(\frac{d \sqrt{2}}{4}\right)^{2}+\left(\frac{d}{2}\right)^{2}}=\frac{d}{4} \sqrt{6}$. Chord $M K$ can be expressed as the hypotenuse of triangle $M O K: M K=\frac{d \sqrt{2}}{2}$, so $M E=E K=\frac{d \sqrt{2}}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6c6a480aadd48812424cg-2.jpg?height=594&width=629&top_left_y=908&top_left_x=288)

The length of segment $A E$ can be found using the theorem of intersecting chords: $A E \cdot B E=M E \cdot E K$, so $A E=\frac{M E \cdot E K}{B E}=\frac{d \sqrt{6}}{12}$, so the ratio of the lengths of the segments: $A E: B E=\frac{d \sqrt{6}}{12}: \frac{d \sqrt{6}}{4}=1: 3$.

Find the side $A B$ of rectangle $A B C D$ as the sum of the lengths of its components: $A B=A E+B E=\frac{d \sqrt{6}}{12}+\frac{d \sqrt{6}}{4}=\frac{d \sqrt{6}}{3}$. Right triangles $B O E$ and $B C D$ are similar by two angles: besides the right angles, they also have $\angle 1=\angle 2$ - alternate interior angles when lines $A B$ and $C D$ are parallel. Therefore, the sides of the triangles are proportional, set up the proportion $\frac{B E}{B D}=\frac{E Q}{B C}$, so $B C=\frac{E O \cdot B D}{B E}=\frac{d}{\sqrt{3}}$. Now calculate the area of rectangle $A B C D$:

$$
S_{A B C D}=A B \cdot B C=\frac{d \sqrt{6}}{3} \cdot \frac{d}{\sqrt{3}}=\frac{d^{2} \sqrt{2}}{3}
$$

Answer: $\frac{d^{2} \sqrt{2}}{3}$.