4. Circle $\omega$ with radius 8 and center $O$ is inscribed in an acute-angled triangle $E N T$ and touches its sides $E N$ and $N T$ at points $C$ and $A$ respectively. Circle $\Omega$ with radius $\frac{\sqrt{145}}{2}$ and center $B$ is circumscribed around triangle $A C N$.

a) Find $O N$.

b) Suppose it is additionally known that the ratio of the area of triangle $B T E$ to the area of triangle $E N T$ is $\frac{7}{10}$. Find the length of the angle bisector $N L$ of triangle $E N T$, as well as its area.
Answer: a) $O N=\sqrt{145}$, b) $N L=\frac{5 \sqrt{145}}{3}, S=360$.

Solution. a) The radius drawn to the point of tangency is perpendicular to the tangent, so angles $O C N$ and $O A N$ are right angles, i.e., the segment $O N$ is seen from points $A$ and $C$ at a right angle. Therefore, the circle constructed on the segment $O N$ as a diameter also passes through points $A$ and $C$, meaning this is the circle $\Omega$. The segment $O N$ is equal to twice the radius of this circle, i.e., $O N=\sqrt{145}$.

b) Let the point of tangency of the circle with side $E T$ be $F$, and the height of the triangle from vertex $N$ be $N H$. Let $\angle L N H=\gamma$. The center of the inscribed circle is the point of intersection of the angle bisectors of the triangle, so $O \in N L$. From triangle $C N O$, we find that $C N=\sqrt{O N^{2}-O C^{2}}=9$.

Since triangles $B T E$ and $E N T$ share the base $E T$, their areas are proportional to the heights drawn to this base, and the ratio of these heights is $N L: B L$. From this, we get $\frac{7}{10}=\frac{B L}{N L}=\frac{N L-B N}{N L}$, $7 N L=10 N L-10 B N, N L=\frac{10}{3} B N=\frac{10}{3} \cdot \frac{\sqrt{145}}{2}=\frac{5 \sqrt{145}}{3}, L O=N L-O N=\frac{2 \sqrt{145}}{3}$.

Since $O F \| N H, \angle L O F=\angle L N H=\gamma$; from triangle $L O F$, we get $\cos \gamma=\frac{F O}{L O}=\frac{12}{\sqrt{145}}$. From triangle $N L H$, we find that $N H=N L \cos \gamma=20$.

We can express the area of triangle $E N T$ in two ways. On one hand, $S_{E N T}=\frac{1}{2} \cdot E T \cdot N H=10 \cdot E T$. On the other hand, $S_{E N T}=p r$, where $r=8$ is the radius of the inscribed circle, and $p$ is the semiperimeter of the triangle. Since $T F=T A, E F=E C, N C=N A$, we get $p=T F+E F+C N=E T+C N=E T+9, S_{E N T}=8(E T+9)$. We obtain the equation $10 \cdot E T=8(E T+9)$, from which $E T=36, S_{C M F}=10 \cdot E T=360$.