Suppose that $A$, $B$ and $C$ are non-zero vectors in $\mathbb R^2$. Show that if $A$ and $B$ are orthogonal and $B$ and $C$ are orthogonal then $A$ and $C$ are parallel.
I feel like this should be very simple. My initial thought was to do a proof by contradiction. So if we assume that $A$ is no parallel to $C$, then we end up with two different $B$'s, one orthogonal to $A$ and one to $C$, but I don't think that is rigorous enough.
Since $A \perp B$, we have $B^TA=0$ implying $A$ and $B$ as two linearly independent vectors of $R^2$. Thus $C=\lambda_1A+\lambda_2B$ for some $\lambda_1,\lambda_2\in R$. Since $B\perp C$, $B^TC=0$
But $$B^TC=\lambda_1B^TA+\lambda_2B^TB$$ Hence $$\lambda_2B^TB=0$$ i.e. $\lambda_2=0$ (since $B\ne0$). So $$C=\lambda_1A$$ implying $$A \parallel C$$