jmc

We know that the perimeters of the two small triangles are $3+CD+AD$ and $4+BD+AD$. Setting both equal and using $BD+CD=5$, we have $BD=2$ and $CD=3$. Now, we simply have to find the area of $△ABD$. Since $\frac{BD}{CD}=\frac{2}{3}$, we must have $\frac{[ABD]}{[ACD]}=2/3$. Combining this with the fact that $[ABC]=[ABD]+[ACD]=\frac{3\cdot 4}{2}=6$, we get $[ABD]=\frac{2}{5}[ABC]=\frac{2}{5}\cdot 6=\boxed{\frac{12}{5}}$.