Find the intersection of an angle bisector and the opposite side

Question

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:

1. Find the equation of the straight line for the line segment $AD$ that bisects $\angle$$BAC$.
2. Find the equation of the straight line that passes through points $B$ and $C$.
3. 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!

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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 (=

Answer 1

Alternatively, using the Angle Bisector Theorem, we have $BA:AC = BD:DC$.

The lengths of $BA$ and $AC$ are easy to find; finally use the section formula on $BC$.

Answer 2

Whoops, this is an elaboration on the other answer.

Here's one way this can be done. Thinking about these like vectors, if we let $t:=\frac{BD}{CD},$ we can say $$D=tB+(1-t)C.\tag{$*$}$$ We can compute $t$ by the angle bisector theorem, saying $$t=\frac{AB}{AC}.$$ We can compute the lengths $AB$ and $AC$ from the coordinates, from which $(*)$ will finish.

By the way, this is a bit more elegant with barycentric coordinates instead of Cartesian ones.

Answer 3

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.