Rectangle Inscribed in an Isosceles Triangle

$AB=3x$ and $AC=2x$

Let point $H$ be the midpoint of $AC$ , then $AH=x$ . By pythagorean theorem,

$(BH)^2=(AB)^2-(AH)^2=(3x)^2-x^2=8x^2$

$BH=\sqrt{8x^2}=2x\sqrt{2}$

$AG=FC=\dfrac{2x-8}{2}=x-4$

Since $\triangle ADG \sim \triangle ABH$ , we have

$\dfrac{DG}{AG}=\dfrac{BH}{AH}$

$\dfrac{DG}{x-4}=\dfrac{2x\sqrt{2}}{x}$

$DG=2\sqrt{2}(x-4)=2x\sqrt{2}-8\sqrt{2}$

So the area of the rectangle is

$8 (2x\sqrt{2}-8\sqrt{2})=\boxed{16x\sqrt{2}-64\sqrt{2}}$

Since $\overline{AC}=2x$ , $\overline{GF}=8$ , and $\overline{AG} = \overline{FC}$ we can solve for side lengths $\overline{AG}$ and $\overline{FC}$ in terms of $x$ .

Let $\overline{AG}$ and $\overline{FC}\ = y$

$2y+8 = 2x$ $\Rightarrow$ $y=x-4$

Now, notice that $\overline{DE}\parallel\overline{AC}$ . It follows that $\triangle{DBE}\sim\triangle{ABC}$ . Because of this, the ratio of the corresponding sides can be expressed as such:

Let $\overline{AD} = z$

$\frac{8}{2x} = \frac{3x-z}{3x}$ $\Rightarrow$ $z=3x-12$ .

Now, observe that $\frac{\overline{AD}}{3} = \overline{AG}$ . Using this information along with the Pythagorean theorem, we can solve for side length $\overline{DG}$ as follows:

Let $\overline{AG}=a$ , $\overline{DG}=b$ , and $\overline{AD}=c$

$a^{2}+b^{2}=c^{2}$ $\Rightarrow$ $a^{2}+b^{2}=(3a)^{2}$ $\Rightarrow$ $b=a\sqrt{8}$

Now that we have $\overline{DG}$ in terms of $a$ , it is possible to find it in terms of $x$ as such:

$\overline{DG}=(x-4)\sqrt{8}$ $\Rightarrow$ $\overline{DG}=2x\sqrt{2}-8\sqrt{2}$

Now that we have $\overline{DG}$ in terms of $x$ the area of rectangle $DEFG$ in terms of $x$ is as follows:

Area = $8 \times{(2x\sqrt{2}-8\sqrt{2})}$ $\Rightarrow$ $\boxed{16x\sqrt{2}-64\sqrt{2}}$