Confused about a step in a proof that $a\times (b \times c) = (a\cdot c)b - (b\cdot c)a$

Let $a, b, c \in \mathbb{R}^3$. I'm trying to understand a step in a proof of the result

$$a\times (b \times c) = (a\cdot c)b - (b\cdot c)a.$$

The author begins with $a\times (b \times c) = xa+yb$ for some $x,y\in \mathbb{R}$, and then writes

$$ c\cdot(xa+yb) = x(c\cdot a)+y(c\cdot b) = 0.$$

Why is the result equal to zero?

That website's presentation is not too clear and has some minor typos as well; I would suggest wiki as an alternative.

In any case, keeping with your setup, first keep the following property in mind:

RULE 1: The cross product between two vectors is always orthogonal to both vectors.

So first off, going along with your website, in your setup it should be $(a\times b)\times c=xa+yb$ for some scalars $x,y$; this is not the same as what you have since the cross product is not associative.

Ok, so why can we write the triple product $(a\times b)\times c$ as a linear combo of $a$ and $b$? By rule 1, this triple product must be orthogonal to $a\times b$; since $a\times b$ is itself orthogonal to $a,b$ by rule 1, then this triple product is coplanar to $a,b$ allowing us to write it as a linear combo of the two.

By rule 1, the triple product is also orthogonal to $c$ and this gives the zero result that you are confused about.