Prove that barycentric coordinates is the ratio of subtriangles to the whole triangle.

geometrytrigonometry

Prove that each of the barycentric coordinates $w_i$ of a point P lying inside the triangle is given by the ratio of the area of the subtriangle $U_i$ formed using P and the two vertices $P_{i+1}$(mod3) and $P_{i+2}$(mod3) to the area of the triangle T.

I know this is a pretty common theorem (the areal coordinates), but could not find a friendly proof about it. Thanks in advance.

The image of triangle

Best Answer

First of all one can prove that barycentric coordinates are invariant under an affine transformation. Indeed, let $\lambda_1$, $\lambda_2$, $\lambda_0$ be the barycentric coordinates of a point $\mathbf P$: $$ \mathbf P=\lambda_1\mathbf P_1+\lambda_2\mathbf P_2+\lambda_0\mathbf P_0,\quad\text{with}\quad \lambda_1+\lambda_2+\lambda_0=1. $$ Under an affine transformation the vertices of the reference triangle change to: $$ \mathbf P_1'=M\mathbf P_1+\mathbf b,\quad \mathbf P_2'=M\mathbf P_2+\mathbf b,\quad \mathbf P_0'=M\mathbf P_0+\mathbf b, $$ where $M$ is a linear transformation and $\mathbf b$ a vector. The point with the same barycentric coordinates of $\mathbf P$ with respect to the transformed vertices is $$ \lambda_1\mathbf P_1'+\lambda_2\mathbf P_2'+\lambda_0\mathbf P_0' =M(\lambda_1\mathbf P_1+\lambda_2\mathbf P_2+\lambda_0\mathbf P_0)+(\lambda_1+\lambda_2+\lambda_0)\mathbf b=M\mathbf P+\mathbf b=\mathbf P', $$ which is what we wanted to prove.

We can now prove the main theorem for a particular triangle and it will hold for any triangle, because any triangle can be transformed into any other triangle by an affine transformation, and ratios of areas are invariant.

Take then this particular triangle: $$ \mathbf P_1=(1,0),\quad \mathbf P_2=(0,1),\quad \mathbf P_0=(0,0). $$ If $\mathbf P=(x,y)$ is any point in the plane we can write: $$ \mathbf P=x\mathbf P_1+y\mathbf P_2+(1-x-y)\mathbf P_0, $$ so that $(x,y,1-x-y)$ are the barycentric coordinates of $\mathbf P$.

But it is immediate to realise that, in this case, $x$ is the ratio between the signed area of triangle $\mathbf P\mathbf P_2\mathbf P_0$ and the area of the reference triangle, while $y$ is the ratio between the signed area of triangle $\mathbf P\mathbf P_0\mathbf P_1$ and the area of the reference triangle, so our proof is complete.