I'm thinking the answer is choice A but I want someone to back up my reasoning/check. So since DE and DF are perpendicular to sides AB and AC respectively that must make EDFA a rectangle. Therefore AF must be 4.5 and AE must be 7.5. Since they state AB = AC that must mean EB is 4.5 and CF is 7.5. Therefore CF rounded to the nearest whole number is 8
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It's an isosceles triangle so $\triangle BED$ is similar to $\triangle CFD$ (base angles are congruent). So we have $\frac{ED}{FD}=\frac{BD}{DC}$ or $DC=\frac{5}{3}BD$. We also know that $BD+DC=24$ hence $(1+\frac{5}{3})BD=24$, $BD=9$, $DC=15$. Now we just use Phythagor: $CF=\sqrt{15^2-7.5^2}\approx 13 $
Since $\triangle ABC$ is isosceles, $\angle ABC \cong \angle ACB$. Since $\angle BED$ and $\angle CFD$ are right angles, they are congruent. Hence, $\triangle BED \sim \triangle CFD$ by the Angle-Angle Similarity Theorem. Hence, $$\frac{|CD|}{|BD|} = \frac{|CF|}{|BE|} = \frac{|FD|}{|ED|} = \frac{7.5}{4.5} = \frac{5}{3}$$ Since $|CD| + |BD| = 24$, we obtain \begin{align*} |CD| + |BD| & = 24\\ \frac{5}{3}|BD| + |BD| & = 24\\ \frac{8}{3}|BD| & = 24\\ |BD| & = 9 \end{align*} Thus, $$|CD| = \frac{5}{3}|BD| = 15$$ Since $$\frac{|DF|}{|CD|} = \frac{7.5}{15} = \frac{1}{2}$$ we may conclude that $\angle FCD = 30^\circ$ and $\angle FDC = 60^\circ$. Thus, $\triangle CDF$ is a $30^\circ, 60^\circ, 90^\circ$ right triangle. Since the ratio of the side lengths in a $30^\circ, 60^\circ, 90^\circ$ right triangle is $1: \sqrt{3}: 2$, $\overline{DF}$ is opposite the $30^\circ$ angle, and $\overline{CF}$ is opposite the $60^\circ$ angle,
$$|CF| = \sqrt{3}|DF| = 7.5\sqrt{3} \approx 12.99$$ so the best answer is $13$.