The center of the circle inscribed in a quadrilateral lies on its diagonal, which is equal to 5. It is known that the perimeter of the quadrilateral is 14, and the area is 12. Find the second diagonal and the sides of the quadrilateral.
Let $AC=5$ be the diagonal of the given quadrilateral $ABCD$, and $O$ be the center of the inscribed circle in the quadrilateral. The center of the circle inscribed in an angle lies on the bisector of that angle, so triangles $ABC$ and $ADC$ are equal by side and two adjacent angles.

Denote $AB=AD=x, BC=CD=y, \angle ABC=\alpha$. Then

$$
x+y=7, \frac{1}{2} x y \sin \alpha=6, x y=\frac{12}{\sin \alpha}, x^2+y^2-2 x y \cos \alpha=25,
$$

$$
x^2+y^2-2 x y \cos \alpha=(x+y)^2-2 x y-2 x y \cos \alpha=49-2 x y(1+\cos \alpha)=25, x y=\frac{12}{1+\cos \alpha}
$$

From the equality $\frac{12}{1+\cos \alpha}=\frac{12}{\sin \alpha}$, it follows that $1+\cos \alpha=\sin \alpha$. Squaring both sides of this equation and simplifying, we get the equation $\cos \alpha(1+\cos \alpha)=0$, and since $0<\alpha<180^{\circ}$, then $\alpha=90^{\circ}$.

Thus

$$
\left\{\begin{array}{l}
x+y=7 \\
x y=12
\end{array}\right.
$$

From this system, we find that $x=3, y=4$ or $x=4, y=3$, and the area of triangle $ABC$ is 6.

Points $A$ and $C$ are equidistant from the endpoints of segment $BD$, so $AC$ is the perpendicular bisector of segment $BD$. From the equality $\frac{1}{2} AC \cdot BD=12$ (the area of quadrilateral $ABCD$), we find that

$$
BD=\frac{2 \cdot 12}{AC}=\frac{2 \cdot 12}{5}=\frac{24}{5}
$$

## Answer

$\frac{24}{5}, 3,4,3,4$.

## Problem