Trigonometry? Get the "half" of a triangle from hypotenuse and cathetus I've only got the following parts of a triangle:


*

*Line A to B 

*Line B to C


And optionally the Line from A to C if needed?
I'm trying to get the point X

Now the problem is, i've got absolutly no idea of trigonometry so I dont even know how to search for it.
Thanks in advance!
 A: OK, then working off your diagram above and assuming AX is perpendicular to BC, you can proceed as follows:
Let's call the co-ordinates of point A $(x_a, y_a)$, point B $(x_b, y_b)$, point C $(x_c, y_c)$ and point X $(x, y)$.
For line BC we can write: $\frac{y-y_c}{x-x_c}=\frac{y_b-y_c}{x_b-x_c}=m$ where $m$ represents the slope of this line. This leads to:
$y=y_c+mx-mx_c\tag{1}$
Similarly, for line AX we can write: $\frac{y-y_a}{x-x_a}=-\frac{1}{m}$ (since line AX is perpendicular to line BC). This leads to:
$y=y_a-\frac{x}{m}+\frac{x_a}{m}\tag{2}$
Now just equate equations (1) and (2) and solve for $x$ and then use either of these equations to find $y$. $(x, y)$ will then be the point you desire.
You will have to be careful in dealing with vertical lines as their slopes will turn out to be infinite.
To help you along, if you equate (1) and (2) you'll get:
$y_c+mx-mx_c=y_a-\frac{x}{m}+\frac{x_a}{m}$
Then multiply both sides by $m$ to get:
$my_c+m^2x-m^2x_c=my_a-x+x_a$
Take all terms involving $x$ to the left-hand-side of the equals and all others to the right-hand-side to get:
$(m^2+1)x=m(y_a-y_c)+x_a+m^2x_c$
Then divide both sides by $(m^2+1)$ to get:
$\large x=\frac{m(y_a-y_c)+x_a+m^2x_c}{m^2+1}\tag{3}$
This will give you a value for the $x$ co-ordinate of point X which you can then use in equation (1) to calculate the corresponding $y$ value for point X.
A: $X=B+\frac{(A-B)\cdot (C-B)}{||C-B||^2} (C-B)$
