# Number of line segments intersecting diagonals are divided into in a convex polygon

I am currently self-studying introductory combinatorics, and I don't fully get an example in the book.

The question was as follows: If no three diagonals of a convex decagon meet at the same point, into how many line segments are the diagonals divided by their intersections?.

So I understand that the total number of intersections would be $\binom {10} {4} = 210$ since for every 4 vertices there will be 1 intersection, and also the number of diagonals would be $10 \choose 2$-10 = 35. I also understand that the number of line segments are $k+1$ when there are $k$ intersections along a line. I however don't get the answer the author gave which was $35+2\times210$. Why were the number of diagonals added?, why did he multiply the number of intersections by 2?, shouldn't it be 4 since for every intersection there will be 4 segments?. If anyone can explain this to me, I'd be very grateful. Also if someone can provide a different way to solve this problem, that would be awesome.

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