Reasoning for equality of Cauchy Schwarz inequality holds

Question

I am trying to understand the proof of the Cauchy-Schwarz Inequality. I understand that, in addition to that, there is a remark that the equality holds if and only if one of term is scalar multiple of the other. And in the book, this is the proof given.

I know there are many alternative proof on MSE but I wanted to understand argument in book.

It is as follows :
$$\langle x,y\rangle=\Vert x\Vert \,\Vert y\Vert\implies\left\langle\frac {x}{\Vert x\Vert} ,\frac{y}{\Vert y\Vert}\right\rangle=1\implies\frac {x}{||x||}=\frac{y}{||y||}$$ Thus,$$x= \Vert x\Vert\,\frac{y}{\Vert y\Vert}.$$
I don't understand second to last line Any help will be appreciated.
I had associated screenshot of proof

Answer 1

Near the start of the proof, they show that for unit length vectors $$\langle x-y, x-y\rangle = 2 - 2\langle x,y\rangle.$$ But recall that $\langle u,u\rangle = 0 \iff u = 0$. Thus if $\langle x,y\rangle = 1$, then $$\langle x-y, x-y\rangle = 2 - 2 = 0 \implies x -y = 0 \implies x = y$$

Instead of assuming $x$ and $y$ are unit length, if we repeat these calculations with $\frac{x}{\Vert x\Vert}$ and $\frac{y}{\Vert y\Vert}$, you obtain precisely their proof.

Answer 2

That step was explained way before in the proof:

If $\langle x,y\rangle=1$, from the above chain of inequalities we deduce that $\langle x-y,x-y\rangle=0$.

(of course, this in the case $\|x\|=\|y\|=1$, as stated in the book).