In a triangle ABC, D is a point on the side BC and M, N are length of perpendicular dropped on line AD from the vertices B and C respectively. Is M > N?
(I) AB > AC
(II) BD < DC
(I) AB > AC
(II) BD < DC
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IMO CThe Iceman wrote:In a triangle ABC, D is a point on the side BC and M, N are length of perpendicular dropped on line AD from the vertices B and C respectively. Is M > N?
(I) AB > AC
(II) BD < DC
I guess I got my answer...nisagl750 wrote:The Iceman wrote:In a triangle ABC, D is a point on the side BC and M, N are length of perpendicular dropped on line AD from the vertices B and C respectively. Is M > N?
(I) AB > AC
(II) BD < DC
If AD is perpendicular to BC, It is not possible to draw perpendiculars on AD from points B & C.
Am I missing something?
Think again. It indeed is possible even though AD were not a perpendicular to BC.nisagl750 wrote: I guess I got my answer...
AD is INDEED the perpendicular to BC....This is the only possible way
How..?The Iceman wrote:
Think again. It indeed is possible even though AD were not a perpendicular to BC.
Even if <ADB is obtused, a perpendicular can be dropped from B to AD produced to E. In this case BE is the perpendicular and E lies outside the triangle.nisagl750 wrote: How..?
What am I missing?
Thats called thinking outside the Triangle.....The Iceman wrote:
Even if <ADB is obtused, a perpendicular can be dropped from B to AD produced to E. In this case BE is the perpendicular and E lies outside the triangle.
OA is B.nisagl750 wrote:Thats called thinking outside the Triangle.....The Iceman wrote:
Even if <ADB is obtused, a perpendicular can be dropped from B to AD produced to E. In this case BE is the perpendicular and E lies outside the triangle.
So, what is the OA?
Let D' be the midpoint of BC and let X' and Y' be the feet of the perpendicular from B and C to AD' respectively => X' = Y'. As D' shifts to right or left, we can know which of M or N is bigger. Thus, statement II answersprat_agl wrote:Can some one please explain how the OA is B?
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