Understanding specific explanation of centre value of 3x3 magic square EDIT. Put another way, we know from the summation that each line must equal 15. 4 lines should equal 60, but there is some over-counting. How can we know that this over-counting is by 15, without knowing in advance that the middle square is worth 5?
I know there are answers on here proving why the centre square of a 3x3 magic square must be 5. However, I would like help understanding a specific explanation because it bothers me that something explained casually on a video as an "of course" does not yet make sense to me.
Based on this image:

the instructor claims that $4S = total sum + 3.centre$.
I think variable names may be causing me some confusion here. What is $totalsum$? I thought originality it meant the sum of the 4 lines, but then the instructor goes on to say that we know $totalsum$ is $3S$ which we know is $45$.
I know that $4S = 60$, and I can see that the centre square is counted 4 times. Beyond this though, the penny is stuck in the slot..
I would be grateful if someone could provide an explanation based on the approach given, but filling in the gaps and clarifying the argument.
The transcript from the lesson is below. Please note that the transcription software has made things sound worse than they actually are, even though for me the explanation was not sufficiently clear to understand in spite of a sincere effort to do so.
And the trick is that we can predict what is in the centre. So look, we have sum up four lines. This is the one line, this is the second line, third line and fourth line. First we know that if we sum all these numbers along these lines, we get four times the constant, this s, which is the sum of each line and each diagonal. So we have 4S. On the other hand, we can see that everything is counted exactly once, except for the centre which is count for four lines, it's count four times. So the centre, there is additional to 3×center. So 4S is the total sum which we know is 3S and 3×center. So the centre is one third, so this four and three, so the difference is S. And the centre is one third of the sum on every line. In our case is five. So, here is the equation and here is the solution in general, and in our case it's five.
 A: This should be a cleaned-up version of the given explanation. Let the square be
$$\begin{bmatrix}
a&b&c\\d&e&f\\g&h&i\end{bmatrix}$$
Then $S$ is the sum of any row or column or diagonal. We are using $1$ to $9$ exactly once, and the three rows use each cell exactly once, so
$$(a+b+c)+(d+e+f)+(g+h+i)=3S=1+\cdots+9=45\implies S=15\tag1$$
Now we take the sum of all lines passing through the centre:
$$(\color{blue}a+\color{blue}e+\color{blue}i)+(\color{blue}b+e+\color{blue}h)+\color{blue}c+e+\color{blue}g)+(\color{blue}d+e+\color{blue}f)=4S\tag2$$
The blue variables are precisely the three-line sum $(1)$ we have already worked out. Thus $(2)=4S$ equals $3S+3e$, i.e. $S=3e$ and $e=5$.
A: The "total sum" is $T=1+2+ ... +9=45.$
Each of the 3 rows has the same sum, and the sum of all the members of all 3 rows is $T.$ So the sum of each row is $T/3=15.$ Each diagonal and each column also has this sum $15.$
Let $C$ be the number in the center. Each of the 4 green lines sums to $15.$ So adding these sums gives a value $\quad  4$x$15=60.$ But this adds each non-center cell once, plus $4$x$C.$
This is the same as adding $every$ cell once (giving $T$), plus $3$x$C.$
That is, the sum of the sums of the 4 green lines is $T+3$x$C.$
Therefore $60=T+3$x$C=45+3$x$C.$
A: I appreciate everyone's input. It has helped me to identify the crux of the issue I had understanding the explanation. This was that it was not explicitly pointed out that the 4 lines form a covering of the whole grid with some over-counting, and thus knowing the actual sum for the whole grid makes it easy to represent the problem as $60 = 45 + 3x$.
This point about the 4.lines covering the whole grid is probably so obvious to someone who already understands the explanation that it doesn't seem to be worth mentioning, yet for me the entire explanation hinged on it. One thing that would have helped is a better variable name: "total sum" seemed to me to mean the same as $4S$. I think "grid sum" or similar would have made the situation much clearer.
