4. Given an isosceles trapezoid \(ABCD (AD \parallel BC, AD > BC)\). A circle \(\Omega\) is inscribed in angle \(BAD\), touches segment \(BC\) at point \(C\), and intersects \(CD\) again at point \(E\) such that \(CE = 7\), \(ED = 9\). Find the radius of the circle \(\Omega\) and the area of trapezoid \(ABCD\).
Answer: $R=2 \sqrt{7}, S_{A B C D}=56 \sqrt{7}$.

Solution. Let the points of tangency of the circle with the sides $A B$ and $A D$ of the trapezoid be denoted as $K$ and $W$ respectively. By the tangent-secant theorem, $D W^{2}=D E \cdot D C=9 \cdot 16, D W=12$. Since $C$ and $W$ are points of tangency of the circle with parallel lines $B C$ and $A D$, the segment $C W$ is the diameter of the circle, perpendicular to these lines. By the Pythagorean theorem in the right triangle $C D W$, we find that $C W=\sqrt{16^{2}-12^{2}}=4 \sqrt{7}$. Therefore, the radius $R$ of the circle is $\frac{1}{2} C W=2 \sqrt{7}$. Let $B C=x$. Then $B K=x$ (tangents drawn from a point to a circle are equal), and since the trapezoid is isosceles, $A K=A B-B K=16-x$. Thus, $A W=A K=16-x$. From this, we get that the sum of the bases is $B C+A D=x+(28-x)=28$, and the area of the trapezoid is $\frac{1}{2} \cdot 4 \sqrt{7} \cdot 28=56 \sqrt{7}$.