Triangle $ABC$ has side lengths $AB = 11, BC=24$, and $CA = 20$. The bisector of $\angle{BAC}$ intersects $\overline{BC}$ in point $D$, and intersects the circumcircle of $\triangle{ABC}$ in point $E \ne A$. The circumcircle of $\triangle{BED}$ intersects the line $AB$ in points $B$ and $F \ne B$. What is $CF$?
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$
[2021AMC12BFallP24.png](https://artofproblemsolving.com/wiki/index.php/File:2021AMC12BFallP24.png)
By the Law of Cosine $\cos A = \frac{AC^2 + AB^2 - BC^2}{ 2 \cdot AC \cdot AB} = \frac{20^2 + 11^2 - 24^2}{2\cdot20\cdot11} = -\frac18$
As $ABEC$ is a cyclic quadrilateral, $\angle CEA = \angle CBA$. As $BDEF$ is a cyclic quadrilateral, $\angle CBA = \angle FEA$.
$\because \quad \angle CEA = \angle FEA \quad \text{and} \quad \angle CAE = \angle FAE$
$\therefore \quad \triangle AFE \cong \triangle ACE$ by $ASA$
Hence, $AF = AC = 20$
By the Law of Cosine $CF = \sqrt{20^2 + 20^2 - 2 \cdot 20 \cdot 20 (-\frac18)} = \sqrt{900} = \boxed{\textbf{C}~\text{30}}$
Note that $F$ is $C$'s reflection over line $AE$, quadrilateral $ACEF$ is a kite symmetrical by line $AE$, $AE \perp CF$.
~[isabelchen](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Isabelchen)