find the external bisector of a given triangle - Maple Help

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geometry[ExternalBisector] - find the external bisector of a given triangle

 Calling Sequence ExternalBisector(bA, A, ABC)

Parameters

 bA - A-external-bisector of ABC A - vertex of ABC ABC - triangle

Description

 • 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.

Examples

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

define the external bisector bA

 > $\mathrm{ExternalBisector}\left(\mathrm{bA},A,\mathrm{ABC}\right)$
 ${\mathrm{bA}}$ (1)
 > $\mathrm{detail}\left(\mathrm{bA}\right)$
 assume that the names of the horizontal and vertical axes are _x and _y, respectively
 $\begin{array}{ll}{\text{name of the object}}& {\mathrm{bA}}\\ {\text{form of the object}}& {\mathrm{line2d}}\\ {\text{equation of the line}}& \left({2}{}\sqrt{{10}}{+}\sqrt{{4}}\right){}{\mathrm{_x}}{+}{3}{}\sqrt{{4}}{}{\mathrm{_y}}{=}{0}\end{array}$ (2)
 > $\mathrm{bisector}\left(\mathrm{ibA},A,\mathrm{ABC}\right):$
 > $\mathrm{ArePerpendicular}\left(\mathrm{bA},\mathrm{ibA}\right)$
 ${\mathrm{true}}$ (3)