$ABC$ is a right-angled triangle at $A$. $AB=3cm$, $BC=5cm$, $CD=1cm$. If $BE=EC$, then what is the area of the shaded region?
I could solve some parts of this question but got stuck and was able to find the following:
$AC=4cm$, $AD=3cm$
I also did two constructions. They were:
Drawing a line $DG \parallel AB$ and $HE \parallel AB$. $HE$ intersects $BD$ at $O$. Through these and using similarity, I found the area of $\Delta DCG=\frac{3}{8} cm^2$ and quadrilateral $DGEO=\frac{5}{8} cm^2$.
This question was on an olympiad website. I am unable to find the area of $\Delta OFE$. Can someone please help me out? It would be very helpful. Thank you.
Best Answer
Assign coordinates so $A$ is the origin, $B=(3,0)$ and $C=(0,4)$. Then $F=(3t,4t)$ satisfies $3t+4t=3$ or $F=(9/7,12/7)$.
The shoelace formula can now be used to find the shaded area, since $E=(3/2,2)$: $$\frac12\begin{vmatrix} 0&3\\ 9/7&12/7\\ 3/2&2\\ 0&4\end{vmatrix}=\frac12(18/7+6-27/7-18/7)=\frac{15}{14}$$