imc

$AC$ intersects $BC$ at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so $△ABC\sim △BED$. The hypotenuse of right triangle BED is $\sqrt{1^2+2^2}=\sqrt{5}$. $$\frac{AC}{BC}=\frac{BD}{ED}\Rightarrow \frac{AC}{BC}=\frac{2}{1}\Rightarrow AC=2BC$$ $$\frac{AC}{AB}=\frac{BD}{BE}\Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}}\Rightarrow AC=\frac{2}{\sqrt{5}}$$ Since $AC=2BC$, $BC=\frac{1}{\sqrt{5}}$. $△ABC$ is a right triangle so the area is just $\frac{1}{2}\cdot AC\cdot BC=\frac{1}{2}\cdot \frac{2}{\sqrt{5}}\cdot \frac{1}{\sqrt{5}}=\boxed{\frac{1}{5}}$