A quadrilateral $A B C D$ is inscribed in a circle with center $O$. It is known that the diagonals $A C$ and $B D$ are perpendicular. On each side, we construct semicircles externally, as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_42ab6d7ce4e674d231dcg-12.jpg?height=764&width=729&top_left_y=1326&top_left_x=545)

a) Show that triangles $A O B$ and $C O D$ have the same area.

b) If $A C=8 \mathrm{~cm}$ and $B D=6 \mathrm{~cm}$, determine the area of the shaded region.
Solution

a) Initially, observe that

$$
\begin{aligned}
A \hat{O} B + C \hat{O} D & = \\
2 \cdot (A \hat{C} B + C \hat{B} D) & = \\
2 \cdot 90^{\circ} & = 180^{\circ}
\end{aligned}
$$

Let $M$ and $N$ be the midpoints of the sides $AB$ and $CD$, respectively. Then $B O \hat{O} M = \frac{A \hat{O} B}{2} = \frac{180^{\circ} - C \hat{O} D}{2} = C \hat{O} N$. This allows us to conclude that the right triangles $\triangle BMO$ and $\triangle CNO$ have the same angles. Furthermore, we can conclude that they are congruent because they have the same hypotenuse. Therefore, they have the same area, and consequently, the same applies to $\triangle AOB$ and $\triangle COD$.

![](https://cdn.mathpix.com/cropped/2024_05_01_42ab6d7ce4e674d231dcg-13.jpg?height=767&width=741&top_left_y=1044&top_left_x=761)

b) Let $r_{1} = \frac{AB}{2}$ and $r_{2} = \frac{CD}{2}$ be the radii of the semicircles on the sides $AB$ and $CD$, respectively. By the congruence of triangles $BMO$ and $ONC$, we have $ON = BM = r_{1}$. Thus,

$$
r_{1}^{2} + r_{2}^{2} = R^{2}
$$

Similarly, if $r_{3}$ and $r_{4}$ are the radii of the semicircles on $AD$ and $BC$, we have

$$
r_{3}^{2} + r_{4}^{2} = R^{2}
$$

Now, adding the areas of the four semicircles, we get:

$$
\frac{1}{2} \cdot \pi \cdot (r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r_{4}^{2}) = \pi \cdot R^{2}
$$

Thus, the four semicircles completely fill the circle. Subtracting the areas of the circular segments, we conclude that the shaded area is equal to the area of the quadrilateral. Since the diagonals are perpendicular, this will be equal to

$$
\frac{AC \cdot BD}{2} = \frac{8 \cdot 6}{2} = 24 \text{ cm}^{2}
$$