In the diagram, $\triangle ABC$ is right-angled at $C$. Also, points $M$, $N$ and $P$ are the midpoints of sides $BC$, $AC$ and $AB$, respectively.  If the area of $\triangle APN$ is $2\mbox{ cm}^2$, then what is the area, in square centimeters, of $\triangle ABC$? [asy]
size(6cm);
import olympiad;
pair c = (0, 0); pair a = (0, 4); pair n = (0, 2); pair m = (4, 0); pair b = (8, 0); pair p = n + m;
draw(a--b--c--cycle); draw(n--p--m);
draw(rightanglemark(a,c,b));
label("$C$", c, SW); label("$A$", a, N); label("$N$", n, W); label("$M$", m, S); label("$B$", b, E); label("$P$", p, NE);
[/asy]
Since the ratio $AN:AC$ equals the ratio $AP:AB$ (each is $1:2$) and $\angle A$ is common in $\triangle APN$ and $\triangle ABC$, then $\triangle APN$ is similar to $\triangle ABC$.

Since the ratio of side lengths between these two triangles is $1:2$, then the ratio of areas is $1:2^2=1:4$.

Thus, the area of $\triangle ABC$ is $4 \times 2 = \boxed{8}\mbox{ cm}^2$.