The bisector bA of the angle at A of the triangle ABC is a line segment (or its extension) from vertex A that bisects an angle at A.

If the optional argument P is given, the object returned is a line segment AP where P is the intersection of the bisector at A and the opposite sides.

For a detailed description of the bisector bA, use the routine detail (i.e., detail(bA))

Note that the routine only works if the vertices of the triangle are known.

The command with(geometry,bisector) allows the use of the abbreviated form of this command.

$\mathrm{with}\left(\mathrm{geometry}\right)\:$

$\mathrm{triangle}\left(\mathrm{ABC}\,\left[\mathrm{point}\left(A\,0\,0\right)\,\mathrm{point}\left(B\,2\,0\right)\,\mathrm{point}\left(C\,1\,3\right)\right]\right)\:$

$\mathrm{bisector}\left(\mathrm{bA}\,A\,\mathrm{ABC}\right)$

$\mathrm{bA}$

$\mathrm{detail}\left(\mathrm{bA}\right)$

assume that the names of the horizontal and vertical axes are _x and _y, respectively

$\begin{array}{ll}\text{name of the object}& \mathrm{bA}\\ \text{form of the object}& \mathrm{line2d}\\ \text{equation of the line}& -3\mathrm{\_x}\sqrt{4}+\mathrm{\_y}\left(2\sqrt{10}+\sqrt{4}\right)=0\end{array}$

$\mathrm{bisector}\left(\mathrm{bA}\,A\,\mathrm{ABC}\,n\right)$

$\mathrm{bA}$

$\mathrm{detail}\left(\mathrm{bA}\right)$

$\begin{array}{ll}\text{name of the object}& \mathrm{bA}\\ \text{form of the object}& \mathrm{segment2d}\\ \text{the two ends of the segment}& \left[\left[0\,0\right]\,\left[\frac{2\left(\sqrt{10}+1\right)}{2+\sqrt{10}}\,\frac{6}{2+\sqrt{10}}\right]\right]\end{array}$