Knowing the 3 points of a triangle and a Point inside the triangle, how to find the point intersecting the line from a corner to the oposite side? Here is the visualisation :

I Do have the float3 ( x, y ,z coordinates )  representing A,B,C & D
I'm trying to find F that is the intersection starting from B and continuing after D to hit AC in F
( or any other combination for that matter )
Just don't know what formula to apply in this context.
Thanks for your time !
 A: Note that the line in 3d passing through $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ has Cartesian equation $$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$$ OR $$\frac{x-x_2}{x_2-x_1}=\frac{y-y_2}{y_2-y_1}=\frac{z-z_2}{z_2-z_1}.$$
STEP 1: Find the equation of the line passing through A,C and the line passing through B,D by the above formula.
STEP 2: Solve (find the point of intersection) of the two lines you found. We’re talking about straight, infinite lines here and not just line segments, so yes, they intersect. To do this,
Let  the first line be $$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}=r.$$ Let  the second line be $$\frac{x-x_3}{x_3-x_4}=\frac{y-y_3}{y_3-y_4}=\frac{z-z_3}{z_3-z_4}=s.$$ Now find (x,y,z) in terms of r and s. Eg. from the first equation $$x=x_1+(x_2-x_1)r$$ Now, equate the two ordered pairs of (x,y,z) that you’ve just got. You’ll get three equations in two variables so you can solve it easily.
Substitute the value of r in the first to get F(x,y,z).
A: Let's work in $3D$.  You have the coordinates of $A,B,C,D$ in $3D$.  We have to assume that $D$ lies in the plane containing $A,B,C$, i.e. we have to assume that there exists two scalars $t_1, t_2$ such that
$D = B + t_1 BA + t_2 BC $
The green line passing through $B$ and $D$ is given parametrically by
$ \ell : p(t) = B + t BD = B + t ( t_1 BA + t_2 BC ) $
Points of the line $AC$ are of the form
$ Q(t) = (1 - s) A + s C = (1 - s) (B + BA) + s (B + BC) = B + (1 - s) BA + s BC $
therefore, we must have
$ t (t_1 +  t_2) = 1 $
i.e.
$ t = \dfrac{1}{t_1 + t_2} $
From which it follows immediately that
$ F = B + \dfrac{ t_1}{t_1 + t_2 } BA + \dfrac{t_2}{t_1 + t_2} BC $
To re-cap,  given the point $D$, find the scalars $t_1$ and $t_2$ by solving
$ D - B = t_1 BA + t_2 BC $
which is a linear system of three equations in the two unknowns $t_1, t_2$.
Then apply the above formula for $F$.
