Given three non-colinear points on the plane, prove that any point on the plane can be uniquely represented as an affine combination of them (this is barycentric coordinates).
My proof is below. Please verify, critique, or comment. Note: Many proofs are available; this question is to critique my proof.
Let $a,b,c,v \in \mathbb R^2$ with $a,b,c$ non-colinear. Homogenize $a,b,c,v$ to $\tilde a,\tilde b,\tilde c,\tilde v$ by adding a third dimension with entry $1$.
Now $\{\tilde a, \tilde b, \tilde c\}$ are linearly independent, since if $\tilde b = k \tilde a$, then $k = 1$ and $b=a$; and if $\tilde c = k \tilde a + j \tilde b$, then $k = 1 - j$ and $c = a + j(b-a)$. Thus the matrix $A = \begin{bmatrix}\tilde a & \tilde b & \tilde c\end{bmatrix}$ is invertible, and there exists a unique solution to $$A \begin{bmatrix} r \\ s \\ t \end{bmatrix} = \tilde v.$$
Update
If there's something unclear about the question, or that makes it difficult to respond, please let me know and I'll revise. I say this because I've yet to receive any responses, either verifying or pointing out an error.
Well, to be perfectly pedantic, it remains to show that the linear system is equivalent to the problem at hand, which is that $ra+sb+tc = v$ with $r+s+t=1$. To be coherent with the level of detail in showing the augmented vectors remain free, I think it only remains to say, at the very end:
"This linear system is equivalent to the equations $ra+sb+tc = v$ and $r+s+t = 1$."
I was also taught, and find it doesn't hurt clarity, to state what result we're going to show before carrying out the proof. That would be a sentence such as:
"[...] non-colinear. Let us show there exists a unique triplet $(r,s,t)$ summing to $1$ s.t. $ra+sb+tc = v$. "