Let $G$ be the intersection of $AD$, $BE$. Since the centroid divides the medians into ratios $2 : 1$. We get that $AG = \frac{2}{3} AD = 8$, and $BG = \frac{2}{3} BE = 6$. Now, since $\angle AGB$ is a right angle, we know that $AF = BF = GF = \frac{AB}{2}$. By the pythagorean theorem, we have that $AB^2 = AG^2 + BG^2 = 8^2 + 6^2 = 100 \Rightarrow AB = 10 \Rightarrow GF = 5 \Rightarrow CF = 3 GF = 15$. $\blacksquare$

Let G be the centroid. Since the centroid splits medians into a 2:1 ratio, we know that AG=8, GD=4, BG=6, GE=3. Since the medians are perpendicular, by pythag we know that $BD=DC=2\sqrt{13}$ and $AE=EC=\sqrt{73}.$ Also by pythag we get that $AB=10.$ Now by stewarts we get GF=15.

AlephG_64 wrote: Let $ABC$ be a triangle and $D,E$ and $F$ the midpoints of sides $BC, AC$ and $BC$. Medians $AD$ and $BE$ are perpendicular, $AD = 12$ and $BE = 9$. What is the value of $CF$? I believe $F$ is the midpoint of $AB$ and not $BC$. We use Cartesian Coordinates. Let the centroid $G$ of $ABC$ be the origin, and $A(8, 0), B(0, 6)$ This way we have $D(-4, 0), E(0, -3)$ by properties of the centroid Now $C$ would be the reflection of $A$ through $E$, which would be $(-8, -6)$. $F$ would be the midpoint of $AB$ which is $(4, 3)$ Therefore the answer is $\sqrt{12^2+9^2}=15$