Another Look at the Shoelace Formula with Cross Products

A Practice Problem for Shoelace Theorem

In right triangle $ABC$, we have $\angle ACB=90^{\circ}$, $AC=2$, and $BC=3$. Medians $AD$ and $BE$ are drawn to sides $BC$ and $AC$, respectively. $AD$ and $BE$ intersect at point $F$. Find the area of $\triangle ABF$.

Source: AoPS