Given a triangle with vertices $A$, $B$ and $C$ whose coordinates are $(x_1, y_1,z_1)$, $(x_2, y_2,z_2)$ and $(x_3, y_3,z_3)$ respectively, how do I determine the coordinates of $D$ such that $AD$ bisects $\angle$$BAC$?
The first idea that came to my mind is as follows:
- Find the equation of the straight line for the line segment $AD$ that bisects $\angle$$BAC$.
- Find the equation of the straight line that passes through points $B$ and $C$.
- Find the intersecting point of these two straight lines.
Is there an alternative method where I can find the coordinates of $D$ directly or just by using the coordinates of $A$, $B$ and $C$? If not, what is the efficient way (computationally) to determine the coordinates of $D$?
Any suggestions or comments are kindly appreciated!
EDIT
It may be true (as suggested by jjagmath and Derive Foiler) that using barycentric instead of cartesian coordinates is elegant but I forgot to mention that what I am looking for is a cartesian solution (=
Best Answer
If $a$, $b$ and $c$ denote the lengths of the sides $BC$, $CA$ and $AB$, then $D$ has barycentric coordinates $(0:b:c)$ in the triangle $ABC$. This means that $D = \frac{b}{b+c} B + \frac{c}{b+c} C$ as vectors.