smc

Note that the triangles $AFH$ and $CEH$ are similar, as they have the same angles. Hence $\frac{AH}{HC}=\frac{AF}{EC}=\frac{2}{3}$. Also, triangles $AGJ$ and $CEJ$ are similar, hence $\frac{AJ}{JC}=\frac{AG}{EC}=\frac{4}{3}$. We can now compute $[EHJ]$ as $[ACD]-[AHD]-[DEH]-[EJC]$. We have: $[ACD]=\frac{[ABCD]}{2}=35$. $[AHD]$ is $2/5$ of $[ACD]$, as these two triangles have the same base $AD$, and $AH$ is $2/5$ of $AC$, therefore also the height from $H$ onto $AD$ is $2/5$ of the height from $C$. Hence $[AHD]=14$. $[HED]$ is $3/10$ of $[ACD]$, as the base $ED$ is $1/2$ of the base $CD$, and the height from $H$ is $3/5$ of the height from $A$. Hence $[HED]=\frac{21}{2}$. $[JEC]$ is $3/14$ of $[ACD]$ for similar reasons, hence $[JEC]=\frac{15}{2}$. Therefore $[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-\frac{21}{2}-\frac{15}{2}=\boxed{3}$.