A bowtie pattern

Firstly, note that F is the ${centroid}$ of triangle ABC, so the area is split into 6 ${equal}$ areas (a proof of this can easily be found online). This means that area[AFN] = $\frac{area[ABC]}{6}$ . Now by joining N to M it is easy to see that triangle NMC is similar to triangle ABC by ${SF}$ $\frac{1}{2}$ ( as CN = $\frac{1}{2}$ AC and C is the point of enlargement). This means that: ${area[NGP]}$ = $\frac{area[ABC]}{2^{2}\times6}$ = $\frac{area[ABC]}{24}$ . Now area[MGQ] = area[NGP] and area[AFN] = area[MFB], so shaded area = area{ABC]( $\frac{1}{3}$ + $\frac{1}{12}$ ). By a repeated argument it is easy to see that the sequence is: $\frac{1}{3}$ + $\frac{1}{12}$ + $\frac{1}{48}$ + ... = $\frac{1}{3}$ $\displaystyle \sum_{i=0}^\infty$ $\frac{1}{4^i}$ = $\frac{1}{3}$ [ $\frac{1}{1-1/4}$ ] = $\frac{4}{9}$ .

$\therefore$ shaded region = $\frac{4}{9}$ $\times$ ${360}$ = $\boxed{160}$