7. (4 points) Given a triangle $ABC$, where $AB=4, \quad BC=4, \quad AC=1$. A bisector is drawn from point $A$, intersecting the circumcircle of this triangle at point $D$. Find the value of $DI$, where $I$ is the incenter of triangle $ABC$.

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