Question

If AP bisects angle BAC and M is any point on AP.prove that the perpendiculars drawn from M to AB and AC are equal

Answer 1

let the perpendiculars from M meet AB &AC at x and y respectively
given
AP bisects <BAC & M is any point on AP
proof
<BAM=CAM            (given AP bisects)
<AXM= <CYM          (90)
AM=AM                    (COMMON SIDE)
So the triangles AXM is cong. to AYM by AAS
hence,XM=YM  (CPCT)
i.e.the perpendiculars are equal