$\left[\begin{array}{l}{[\text { Projections of the bases, sides, or vertices of the trapezoid }]} \\ {[\text { Law of Sines }}\end{array}\right]$

A quadrilateral $ABCD$ is inscribed in a circle of radius $R$, $P$ is the point of intersection of its diagonals, $AB = CD = 5$, $AD > BC$. The height dropped from point $B$ to side $AD$ is 3, and the area of triangle $ADP$ is $\frac{25}{2}$.

Find the sides $AD$, $BC$, and the radius of the circle $R$.
Prove that the given quadrilateral is an isosceles trapezoid. Denote its bases as \( x \) and \( y \). Use the similarity of triangles \( ADP \) and \( CBP \).

## Solution

We will prove that \( AD \parallel BC \). Suppose this is not the case. Draw a line through vertex \( B \) parallel to \( AD \) until it intersects the circle at point \( C_1 \). The arcs enclosed by parallel chords are equal, and therefore the corresponding chords are equal. Thus,
since points \( B, C \), and \( C_1 \) lie on the same side of line \( AD \), point \( C_1 \) coincides with point \( C \). Therefore, \( AD \parallel BC \). Hence, quadrilateral \( ABCD \) is an isosceles trapezoid or a rectangle.

Let \( BK \) be the perpendicular to \( AD \). Since \( BK < AB \), \( ABCD \) is an isosceles trapezoid, and \( AD \) and \( BC \) are its bases.

Denote \( BC = x \) and \( AD = y \). Then

\[
AK = \frac{AD - BC}{2} = \frac{y - x}{2}, \quad AK = \sqrt{AB^2 - BK^2} = 4
\]

Triangle \( ADP \) is similar to triangle \( CBP \) with the similarity ratio \( \frac{AD}{BC} = \frac{y}{x} \). Therefore, the height \( PQ \) of triangle \( ADP \) is

\[
\frac{y}{x + y} \cdot BK = \frac{3y}{x + y}
\]

Thus, the area of this triangle is

\[
\frac{1}{2} AD \cdot PQ = \frac{1}{2} \cdot \frac{3y^2}{x + y} = \frac{25}{2}
\]

Solving the system of equations

\[
\left\{
\begin{aligned}
\frac{y - x}{2} & = 4 \\
\frac{1}{2} \frac{3y^2}{x + y} & = \frac{25}{2}
\end{aligned}
\right.
\]

we find that \( y = 10 \) and \( x = 2 \) or \( y = \frac{20}{3} \) and \( x = -\frac{4}{3} \). Clearly, only the first solution satisfies the problem's conditions. Further, we find:

\[
\begin{gathered}
DK = AD - AK = 6, \quad BD = \sqrt{BK^2 + DK^2} = 3\sqrt{5}, \quad \sin \angle BAD = \frac{BK}{AB} = \frac{3}{5}, \\
R = \frac{BD}{2 \sin \angle BAD} = \frac{3\sqrt{5}}{2 \cdot \frac{3}{5}} = \frac{5\sqrt{5}}{2}
\end{gathered}
\]

## Answer

\( AD = 10, BC = 2; R = \frac{5\sqrt{5}}{2} \).