2021 Fall AMC 12B Problem 24

Points $A,$ $D,$ $E$ are collinear on the bisector, and $A,$ $B,$ $F$ are collinear on line $AB.$ The power of $A$ with respect to the circle through $B,$ $E,$ $D$ gives $AB \cdot AF = AD \cdot AE.$ Since $\angle BAE = \angle DAC$ and $\angle AEB = \angle ACB$ (subtending $AB$), triangles $ABE$ and $ADC$ are similar, so $AD \cdot AE = AB \cdot AC.$ Therefore $AF = AC = 20.$ Place $A = (0, 0),$ $B = (11, 0).$ From $CA = 20,$ $CB = 24,$ point $C = \left(-\tfrac52, \tfrac{\sqrt{1575}}{2}\right).$ Point $F$ lies on ray $AB$ with $AF = 20,$ so $F = (20, 0).$ Then $CF^2 = \left(20 + \tfrac52\right)^2 + \tfrac{1575}{4}$ $= \tfrac{2025}{4} + \tfrac{1575}{4}$ $= 900,$ so $CF = 30.$ Thus, the correct answer is C.