In the projective plane $P(\mathbb{R}^3)=\mathbb{R}P^2$ a triangle $\Delta ABC$ and a point $D$, not on either side of the triangle are given. Denote $P_1=AD\cap BC$ and let $l$ be a changeable line through $P_1$ with $l\neq AD$. Now denote $P_2=l\cap AC, P_3=l\cap AB$ and $P=BP_2\cap AD$. Show that $PP_3$ passes through a fixed point, i.e. by changing the line $l$, $PP_3$ always passes through that point.
I already managed to show that if such a fixed point exists it must lie on $BC$. Take $l=BC$, then $P_2=C,P_3=B$ and $P=P_1$. This means $PP_3=P_1B$ but $P_1\in BC$ so $PP_3=P_1B=BC$ and so the fixed point lies on $BC$. I don't know how to proceed from here. I thought that maybe a projective transformation could set $A=(0:0:0),B=(0:1:0)$ and $C=(1:0:0)$ but then $D$ would still be unknown. Or is a projective transformation such that $D=(1:1:1)$ possible?
Any hints/help is welcome.
Best Answer
There for their composition $\pi := \pi_{P_1}\circ \pi_{B}$ is projective map from $AD\to AB$ which takes $P$ to $P_3$. Now if we want to prove that $PP_3$ goes through fixed point, you have to prove that $\pi$ is also perspective with respect to some other fixed point $S$ and thus $PP_3$ goes through it.
All you have to prove is that $\pi$ takes intersection point of $AB$ and $AD$ i.e. point $A$ to it self (that is characterisation which tells you whether projective map is perspective). That should not be difficult.