The external bisector bA of the angle at A of the triangle ABC is the line perpendicular to the internal bisector of the angle at A.

For a detailed description of the external 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,ExternalBisector) allows the use of the abbreviated form of this command.

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

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

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

$\mathrm{bA}$

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

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

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

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

$\mathrm{ArePerpendicular}\left(\mathrm{bA}\,\mathrm{ibA}\right)$

$\mathrm{true}$