Colouring a Pentagram

First case: CJ and Dan pick edges that share a vertex. In this case, CJ picks another edge from the shared vertex, which Dan must block by taking the third side of the triangle. Then CJ picks the final edge from the shared vertex, and is now threatening to make two different triangles. Dan cannot block both of these.

Numbering vertices, say CJ picks 12 and Dan picks 15. Then CJ picks 13, so Dan must take 23, and then CJ picks 14, and will win on his next turn by either taking 24 or 34.

Second case: CJ and Dan's edges don't share a vertex. WLOG CJ picks 12 and Dan picks 34. Then CJ picks 15, so Dan must take 25, and now CJ picks 13 and will win on his next turn by either taking 23 or 35.

Edit: sorry, I feel like the problem is a bit ambiguous. Are we always connecting two vertices of the pentagon or are we coloring the smaller line segments in between? I assumed the former; if the latter, then the problem is harder than my solution suggests.

Label the vertices A, B, C, D, and E going clockwise from the top vertex. CJ's strategy is start by coloring in one of the diagonals of the pentagon, then coloring in another diagonal that shares a vertex with the first one, regardless of what Dan does. A sample game might begin:

$\begin{array}{c|c|c} \text{Turn} & \text{CJ} & \text{Dan} \\ \hline 1 & AC & BC \\ 2 & AD & \\ \end{array}$

This combination of beginning moves will threaten to form a triangle, so Dan must block it:

$\begin{array}{c|c|c} \text{Turn} & \text{CJ} & \text{Dan} \\ \hline 1 & AC & BC \\ 2 & AD & CD \\ \end{array}$

In doing so, Dan might threaten to form a triangle of his own, so CJ must block it, but after some number of moves, Dan can no longer threaten to form any triangles:

$\begin{array}{c|c|c} \text{Turn} & \text{CJ} & \text{Dan} \\ \hline 1 & AC & BC \\ 2 & AD & CD \\ 3 & BD & AB \end{array}$

After this happens, CJ can clinch the win by choosing a vertex shared by two diagonals and coloring in a side containing that vertex:

$\begin{array}{c|c|c} \text{Turn} & \text{CJ} & \text{Dan} \\ \hline 1 & AC & BC \\ 2 & AD & CD \\ 3 & BD & AB \\ 4 & DE & \\ \end{array}$

This will always threaten to form two triangles - in the sample game, they are $BDE$ and $CDE$ - which Dan cannot block simultaneously. Therefore, CJ will always win the game.

CJ goes first and chooses one of the sides of the inverted little pentagon in the centre.

Then Dan makes his move, he can choose everything he want to fight back CJ's previous move.

Then CJ chooses one of the diagonal's segments linked with the one he had chosen first.

Finally, Dan tries to choose another segment to put back his opponent but with his third and last move CJ will win every single time, provided that he plays first.

In this game whoever plays first is, basically, the winner.

It's the same concept in tic tactics toe. And in mini miny moe For every even number of moves the first player will win

If both players play optimally then whoever goes first will win assuming playing optimally means to win.