Problem while solving this combinatorial geometry puzzle

I'm a math olympiad trainee and received this problem during one of my practice tests. I have seen such problems before but always struggle to solve them. Any tips / solutions to this would be greatly appreciated.

Q) Gerrard drew 100 lines on the plane, with no three lines passing through the same point. These lines divide up the plane in various parts, and some of those parts are triangle shaped. Gerrard claims that he can draw another line that does not pass through any of the existing intersection points, and it intersects atleast 60 triangle shaped parts. Show that Gerrard's claim cannot be true.

Solution 1:

HINT:

1. Let us call a vertex a point on the plane where 2 lines intersect, and let us call an edge $e$ a line segment where each endpoint of $e$ is a vertex. Then the condition that no 3 lines intersect at a single point gives the condition that each edge $e$ is in at most $1$ triangle.

2. The condition that two lines intersect at at most $1$ point gives the condition that the $101$st line $L$ intersects at most $100$ edges.

3. If $L$ intersects a triangle, then $L$ intersepts $2$ edges of the triangle.

So the number of triangles that $L$ can intersept, is [from 1 and 3 above], the number of edges $L$ intersepts, divided by $2$. By 2 above, this is at most $50$.