Answers

2015-04-07T20:30:07+05:30
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
1 5 1