Isosceles Right-angled Triangle

Point $F$ is a centroid of $\triangle ABC$ therefore $FE=\frac{AE}{3}$ therefore $\triangle CFE=\frac{\triangle AEC}{3}$ . $\triangle AEC = \frac{\triangle ABC}{2}=\frac{12 \times 12}{2 \times 2}=36$ $\triangle CFE=\frac{\triangle AEC}{3}=\frac{36}{3}=\boxed{12}$

Note: Medians divide any triangle into 6 smaller triangles of equal areas.

Let $H$ be a point on $\overline{BC}$ such that $\overline{DH} \perp \overline{CB}$ . So we have $\bigtriangleup ABC \sim \bigtriangleup DBH \Rightarrow \overline{CH} = 9\sqrt{2}$ . And finally: $\bigtriangleup HCD \sim \bigtriangleup ECF \Rightarrow \overline{DF}=2\sqrt{2} \Rightarrow AREA_{ECD} = 12$

$tan$ $Φ =$ $\frac{6}{12}$

$Φ = 26.56505118°$

$θ = 45° - 26.56505118° = 18.43494882°$

$CE =$ $\frac{\sqrt{12^2 + 12^2}}{2}=\frac{\sqrt{288}}{2}$

$tan$ $18.43494882° =$ $\frac{FE}{\frac{\sqrt{288}}{2}}$

$FE =$ $2.828427124$

$A_{shaded} = ½(2.828427124 * \frac{\sqrt{288}}{2}) = 12$ $cm^2$

Draw median line from B that goes through F and intersects AC at G. So we divide triangle ABC into 6 smaller triangles with same area. Let the area of this triangle is x. We have 6x=72, x=12. Then the area of the shaded triangle is 12