A square of side length $3$ has two unit squares placed side by side along the left part of its top edge, as shown. Segment $AE$ joins the top-left corner of the left unit square to the bottom-right corner of the right unit square. Segment $BD$ joins the top-right corner of the left unit square to the bottom-left corner of the large square. If the segments intersect at $C$, what is the area of $\triangle ABC$?

Let $H$ be the foot of the perpendicular from $C$ to $\overline{AB}$. Since $C$ lies on $\overline{AE}$ and $AE$ drops $1$ unit for every $2$ units horizontally, $CH=\frac12 AH$. Since $C$ also lies on $\overline{BD}$, whose vertical rise is $4$ units for every $1$ unit horizontally, $CH=4(BH)$. Thus $\frac12 AH=4BH$. With $AH+BH=AB=1$, we get $AH=\frac89$ and $CH=\frac49$. Therefore
\[
[ABC]=\frac12(AB)(CH)=\frac12\cdot1\cdot\frac49=\boxed{\frac29}.
\]