En el siguiente diagrama, sea $ABCD$ un cuadrado y sean $M,N,P$ y $Q$ los puntos medios de sus lados. Demostrar que\n\[S_{A'B'C'D'} = \frac 15 S_{ABCD}.\]\n[asy]\nimport graph; size(200); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttzz = rgb(0,0.2,0.6); pen qqzzff = rgb(0,0.6,1);\ndraw((0,4)--(4,4),qqttzz+linewidth(1.6pt)); draw((4,4)--(4,0),qqttzz+linewidth(1.6pt)); draw((4,0)--(0,0),qqttzz+linewidth(1.6pt)); draw((0,0)--(0,4),qqttzz+linewidth(1.6pt)); draw((0,4)--(2,0),qqzzff+linewidth(1.2pt)); draw((2,4)--(4,0),qqzzff+linewidth(1.2pt)); draw((0,2)--(4,4),qqzzff+linewidth(1.2pt)); draw((0,0)--(4,2),qqzzff+linewidth(1.2pt);\ndot((0,4),ds); label('$A$', (0.07,4.12), NE*lsf); dot((0,0),ds); label('$D$', (-0.27,-0.37), NE*lsf); dot((4,0),ds); label('$C$', (4.14,-0.39), NE*lsf); dot((4,4),ds); label('$B$', (4.08,4.12), NE*lsf); dot((2,4),ds); label('$M$', (2.08,4.12), NE*lsf); dot((4,2),ds); label('$N$', (4.2,1.98), NE*lsf); dot((2,0),ds); label('$P$', (1.99,-0.49), NE*lsf); dot((0,2),ds); label('$Q$', (-0.48,1.9), NE*lsf); dot((0.8,2.4),ds); label('$A'$', (0.81,2.61), NE*lsf); dot((2.4,3.2),ds); label('$B'$', (2.46,3.47), NE*lsf); dot((3.2,1.6),ds); label('$C'$', (3.22,1.9), NE*lsf); dot((1.6,0.8),ds); label('$D'$', (1.14,0.79), NE*lsf); clip((-4.44,-11.2)--(-4.44,6.41)--(16.48,6.41)--(16.48,-11.2)--cycle);\n[/asy] [ $S_{X}$ denota el área de $X.$ ]