In $\mathbb{P}_2(\mathbb{K})$ with $\mathbb{K}\ne\mathbb{Z}_2$,
Given $ABCD$ in a line with ordering $$A(1,0,0),~D(1,-1,0), ~B(0,1,0), ~C(1,1,0)$$ where $D$ is the harmonic conjugate of $C$ respect to $A$ and $B$
I was wondering why $D$ is the conjugate, I have tried the cross-ratio but cannot obtain that: $$(A,B;C,D)=\frac{[AC][BD]}{[AD][BC]}=-1$$
Have a look at the following figure.
Points $A,B,C,D$ having their third coordinates $0$ are in fact points at infinity in the direction indicated by points
$$A_1(1,0,1), \ B_1(0,1,1), \ C_1(1,1,1), \ D_1(1,-1,1)$$
(same first two coordinates as $A,B,C,D$ but third projective coordinate equal to $1$).
You must know that points at infinity can be considered as straight lines, here $OA_1,OB_1,OC_1,OD_1$, constituting what is called a pencil of lines. What do we see ? That $OA$ and $OB$ are bissectors of $OC$ and $OD$.
This is enough: this situation is a characteristic property of harmonic conjugation for a pencil of lines.
But you may prefer a proof using cross ratio. There is a theorem saying that, whatever the pencil of lines, taking any sectional line intersecting the different lines of the pencil in, say $A',B',C',D'$, the value of cross ratio $[A',B';C',D']$ is the same whatever the sectional line (thus, it is called the "cross ratio of the pencil"). Let us take for example sectional line $(S)=C_1D_1$: it is clear that if we take on $(S)$, $A_1$ as the origin and $C_1$ as the point with abscissa 1, the abscissa of $D_1$ is $-1$ and the abscissa of the intersection of $OB_1$ with $(S)$ is point $B$ (at infinity). In such a situation the value of the cross ratio is $-1$ because $\dfrac{[A_1C_1]}{[A_1D_1]} \times\dfrac{[BD_1]}{[BC_1]}=-1 \times 1$ (intuitively: the second fraction is equal to 1 because $B$ is so "far away" that the distance $BD_1$ is the same as the distance $BD_1$).