7. (4 points) Given a triangle $A B C$, where $A B=2, B C=8, A C=8$. From point $B$, a bisector is drawn which intersects the circumcircle of this triangle at point $D$. Find the value of $D I$, where $I$ is the center of the inscribed circle of triangle $A B C$.
Answer: $\frac{16}{3}$.

Solution: According to the trident lemma, $D I=A D$, and by the Law of Sines in triangle $A D B$, $A D=\frac{A B \sin \angle A B D}{\sin \angle A D B}=\frac{A B}{\sin \angle A C B} \sin \angle \frac{A B C}{2}$. Expressing $\frac{A B}{\sin \angle A C B}$ using the Law of Sines in triangle $A B C$, we get $A D=\frac{A C}{\sin \angle A B C} \sin \angle \frac{A B C}{2}=\frac{A C}{2 \cos \angle \frac{A B C}{2}}=\frac{A C}{\sqrt{2 \cos (\angle A B C)+2}}=\frac{8}{\sqrt{2 \cdot \frac{1}{8}+2}}=\frac{8}{\sqrt{\frac{9}{4}}}=\frac{16}{3}$.