MathDB

In the following diagram, let

ABCD

be a square and let

M,N,P

and

Q

be the midpoints of its sides. Prove that

S_{A'B'C'D'} = \frac 15 S_{ABCD}.

[asy] import 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); draw((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)); dot((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); [/asy]
[

S_{X}

denotes area of the

X.

]

3

Problems(2)