11. The answer is $(\mathbf{C})$. Since $C A$ is tangent to $\omega$, we have $C P=\frac{A C^{2}}{C D}=\frac{54^{2}}{108}=27$. Moreover, $\angle D A C=\angle C P A$, but $A D$ is the angle bisector of $\angle B A C$, so $\angle B A D=\angle C P A$. Consequently, the circumcircle of triangle $A B P$ is tangent to the line $A D$, thus $D B=\frac{D A^{2}}{D P}=$ $\frac{63^{2}}{108-27}=49$. Now, by the angle bisector theorem, $\frac{A C}{C D}=\frac{A B}{B D}$, and therefore we can conclude that $A B=\frac{A C \cdot B D}{C D}=\frac{54 \cdot 49}{27}=98$.
11. Let $A B C$ be a triangle and $D$ the foot of the angle bisector from vertex $A$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $D$. Let $P$ be the second intersection of $\omega$ with the line $B C$. Given that $A C=54, A D=63$ and $C P=108$, find $A B$.
(A) 72
(B) $\frac{147}{2}$
(C) 98
(D) 99
(E) 102