A quadrilateral pyramid $S A B C D$ is inscribed in a sphere, the center of which lies in the plane of the base $A B C D$. The diagonals $A C$ and $B D$ of the base intersect at point $H$, and $S H$ is the height of the pyramid. Find the edges $C S$ and $C D$, if $C H=4, A S=3^{\frac{3}{4}}, A D=3, A B=B S$.
Consider the section of the given sphere with center O by the plane ASC (Fig.1). The plane ASC passes through the perpendicular SH to the plane $ABCD$. Therefore, these planes are perpendicular. Hence, the perpendicular dropped from point $O$ to line $AC$ is also a perpendicular to the plane of the section, and therefore passes through the center of the circle of the section. Consequently, AC is the diameter of the circle of the section, and angle ASC is a right angle. Since $SH$ is the height of the right triangle $ASC$ drawn from the vertex of the right angle, then

$$
AS^2 = AH \cdot AC, \text{ or (15) } 2 = AH(AH + 4) \text{. }
$$

From this equation, we find that $AH = \frac{9}{4}$. Therefore,

$$
SC = \sqrt{CH \cdot AC} = \sqrt{4\left(4 + \frac{9}{4}\right)} = \sqrt{16 + 9} = 5
$$

Since $BA = BS$ (by condition) and $OA = OS$ (as radii of the sphere), points B and O are equidistant from the ends of segment $AS$, i.e., they lie in a plane perpendicular to $AS$ and passing through the midpoint of segment $AS$ (Fig.1). Therefore, $BO \perp AS$. Line $AC$ is the orthogonal projection of the inclined AS on the plane $ABCD$. By the theorem of three perpendiculars, $BO \perp AC$. The diameter $BB1$ of the circle circumscribed around quadrilateral $ABCD$ (Fig.3) is perpendicular to the chord $AC$, so $BB1$ bisects this chord. Therefore, triangle $ABC$ is isosceles $(BA = BC)$. From the equality of chords $BA$ and $BC$, it follows that angles $ADB$ and $CDB$ are equal. Thus, $DH$ is the bisector of triangle $ADC$. By the property of the bisector of a triangle,

$$
\frac{CD}{DA} = \frac{CH}{AH}, \text{ or } \frac{CD}{3} = \frac{\frac{4}{9}}{1} = \frac{16}{9}
$$

from which $CD = \frac{16}{3}$.

## Answer

$5 ; \frac{16}{3}$

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