Suppose we have a square ABCD and a point S in the interior of this square.
Under homothety with centre S and ratio of magnification k>1, this square becomes another square A′B′C′D′.
Prove that the sum of the areas of the two quadrilaterals A′ABB′ and C′CDD′ are equal to the sum of the areas of the two quadrilaterals B′BCC′ and D′DAA′.[asy]
unitsize(3 cm);pair[] A, B, C, D;
pair S;A[1] = (0,1);
B[1] = (0,0);
C[1] = (1,0);
D[1] = (1,1);
S = (0.3,0.6);
A[0] = interp(S,A[1],2/3);
B[0] = interp(S,B[1],2/3);
C[0] = interp(S,C[1],2/3);
D[0] = interp(S,D[1],2/3);draw(A[0]--B[0]--C[0]--D[0]--cycle);
draw(A[1]--B[1]--C[1]--D[1]--cycle);
draw(A[1]--S, dashed);
draw(B[1]--S, dashed);
draw(C[1]--S, dashed);
draw(D[1]--S, dashed);dot("A", A[0], N);
dot("B", B[0], SE);
dot("C", C[0], SW);
dot("D", D[0], SE);
dot("A′", A[1], NW);
dot("B′", B[1], SW);
dot("C′", C[1], SE);
dot("D′", D[1], NE);
dot("S", S, dir(270));
[/asy] geometrygeometric transformationhomothetysquare