In rectangle $ABCD$, points $F,G$ lie on $\overline{AB}$ so that $AF=GB=\frac14AB$, and $E$ is the midpoint of $\overline{DC}$. Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$. The area of rectangle $ABCD$ is $75$. Find the area of triangle $EHJ$.

1. $AH:HC=AF:CE=\frac14:\frac12=1:2$, and $AJ:JC=AG:CE=\frac34:\frac12=3:2$, so $HJ=\left(\frac35-\frac13\right)AC=\frac4{15}AC$.
2. $[EAC]=\frac12[ADC]=\frac14[ABCD]$, so $[EHJ]=\frac4{15}\cdot\frac14[ABCD]=\frac1{15}[ABCD]$.
3. With $[ABCD]=75$: $[EHJ]=\boxed{5}$.