consider 4 points in 2D in clockwise order as A,B,C,D forming a quadrilateral.
If the quadrilateral undergoes affine transformation then affine invariant will be preserved which is given by area of triangle ACD/ABC as per http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.109.5013&rep=rep1&type=pdf (see page 3 affine invariant)
My question is - Why can't it be ratio of areas across another diagonal BD i.e area of triangle ADB/BCD ?
Will both ratios be same ?
Thanks.
On
Based on your last comment I think I see what might be causing you some confusion. The claim is that the ratio of areas of the two triangles to either side of a diagonal of a quadrilateral is invariant under (nonsingular) affine transformation. This is a simple consequence of the fact that affine transformations multiply all areas by a constant factor: the ratio of the areas of any two triangles is invariant. Given this, the choice of diagonal is clearly irrelevant. However, this does not mean that the area ratio obtained by using the other diagonal is equal to the first, only that it, too, is invariant under affine transformations. Indeed, the eight possible cyclic labelings of the vertices generate up to four different area ratios.
The moduli space of (labeled) convex plane quadrangles up to affine equivalence is two-dimensional. This can be seen as follows: Given such a quadrangle $Q\subset{\mathbb R}^2$ with vertices $A$, $B$, $C$, $D$ in counterclockwise order there is a unique affine map $T$ with $T(A)=(0,1)$, $T(B)=(0,0)$, and $T(C)=(1,0)$. It follows that $$T(D)=(u,v),\qquad{\rm whereby}\quad u>0,\quad v>0,\quad u+v>1\ ,$$ is uniquely determined; hence the pair $(u,v)$ is an affine invariant of $Q$.
The two diagonals of $Q$ (resp., $T(Q)$) partition $Q$ into a pair of triangles in two different ways, generating for each pair an area ratio $\lambda$, resp., $\mu$. The numbers $\lambda$ and $\mu$ are affine invariants as well. Doing the computations for $T(Q)$ it is easily seen that $u$ and $v$ are uniquely determined by the values of $\lambda$ and $\mu$. It follows that the pair $(\lambda,\mu)$ completely characterizes $Q$ up to affinity. For a parallelogram we have $\lambda=\mu=1$.