MATHS
posted by Judah on .
AX is the bisector of <BAC p is any point on AX. prove that the perpendicular drawn from Pto AB and AC are equal

let pb and pc be the perpendiculars to AB and AC.
Since Ax is the angle bisector, angle XAB = XAC
using similar triangles,
pb/pa = pc/pa
so,
pb=pc