Familiar problem?

Let $BF=x$ while $CE=y$ . We see that $GF=GE+EF=GE+1=GE+CG=CE=y$ .

Note that $HABG$ is a cyclic quadrilateral so $BGF$ is similar to $HAF$ . Therefore:

$\frac{BF}{HF}=\frac{GF}{AF}$

$\frac{BF}{HC+CG+GF}=\frac{GF}{AB+BF}$

$\frac{x}{1+1+y}=\frac{y}{1+x}$

$\frac{x}{2+y}=\frac{y}{1+x}$

$x(1+x)=y(2+y)$

Moreover, considering $BEF$ where $D, A, C$ are collinear, we employ Menelaus' Theorem as follows:

$\frac{AD}{CD}\frac{BF}{AB}\frac{CE}{EF}=1$

$\frac{1}{2}\frac{x}{1}\frac{y}{1}=1$

$xy=2$

$(xy)^{2}=2^{2}$

$x^{2}y^{2}=4$

Considering that $x(1+x)=y(2+y)$ :

$x(1+x)=y(xy+y)$

$x(1+x)=y(y)(x+1)$

$x=y^{1+1}$

$x=y^{2}$

Now substitute this into $x^{2}y^{2}=4$ :

$x^{2}x=4$

$x^{2+1}=4$

$x^3=4$

$x=\sqrt[3]{4}$

As required, $n=3, a=4$ therefore, $n+a=3+4=\boxed{7}$