In $ \mathbb{RP}^2$ given a triangle $A_1A_2A_3$ and a point $P$ not lying on either one of the edges , set points $$B_i = PA_i \cap A_kA_l,\ k,l \not= i \ \forall i$$ next choose points $C_i \in PA_i\ \forall i$ and calculate the cross ratio $(A_i, B_i,P,C_i)=a_i\ \forall i$.
Show that $C_1, C_3, C_3$ are collinear if and only if $a_1+a_2+a_3=2+a_1a_2a_3$.
I appreciate any help, especially hints as i have no good idea how to tackle this problem.
The setup is invariant under projective transformations, so without loss of generality choose $$A_1=[1:0:0]\qquad A_2=[0:1:0]\qquad A_3=[0:0:1]$$ Then with $P=[p_1:p_2:p_3]$ you get $$B_1=[0:p_2:p_3]\qquad B_2=[p_1:0:p_3]\qquad B_3=[p_1:p_2:0]$$ either via cross-product computation or by observing that points on $A_2A_3$ have a zero in the first place, and $B_1=P-p_1A_1$ is exactly the linear combination of $P$ and $A_1$ which satisfies this property. Likewise for other two. Similarly you can assume $C_i=\lambda_i P+\mu_i A_i$ and then compute the cross ratio with respect to the basis $P,A_i$ as
$$a_i=(A_i,B_i;P,C_i)= \left( \begin{bmatrix}0\\1\end{bmatrix},\begin{bmatrix}1\\-p_i\end{bmatrix}; \begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}\lambda_i\\\mu_i\end{bmatrix} \right)=\\ \frac {\begin{vmatrix}0&1\\1&0\end{vmatrix}\cdot \begin{vmatrix}1&\lambda_i\\-p_i&\mu_i\end{vmatrix}} {\begin{vmatrix}0&\lambda_i\\1&\mu_i\end{vmatrix}\cdot \begin{vmatrix}1&1\\-p_i&0\end{vmatrix}}= \frac{\mu_i+\lambda_ip_i}{\lambda_ip_i}\\ \lambda_ip_ia_i=\mu_i+\lambda_ip_i\\ 0=\mu_i+\lambda_ip_i(1-a_i)$$ Solutions to this are as usual only defined up to some scalar factor, but one simple solution would be the following: $$\mu_i=p_i(a_i-1)\qquad \lambda_i=1\\ C_i=P + p_i(a_i-1)A_i $$
All three points are collinear if their determinant vanishes.
$$0=\det[C_1,C_2,C_3]=\begin{vmatrix} p_1+p_1(a_1-1) & p_1 & p_1 \\ p_2 & p_2+p_2(a_2-1) & p_2 \\ p_3 & p_3 & p_3+p_3(a_3-1) \end{vmatrix}$$
The rest is just a bit of algebra.