SAT II Geometry Find the missing side length 
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 
 A: 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 $13$.
A: 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 $ 
A: Here is my thoughts on this one. It is indeed clear that triangles BED and CDF are similar. If $BD = y$ then DC is $24-y$. Using similarity we get the ratio $y/4.5 = (24-y)/7.5$ from which follows $y=9$ and so $CD=15$. With Pythagorean theorem you find 12.99 
