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

IMO C

I have one doubt though.
Since D is a point on side BC of the triangle, D will be somewhere between B & C
i.e. B & C will lie on either side of D (one to the left and one to the right.

If AD is perpendicular to BC, It is not possible to draw perpendiculars on AD from points B & C.
If ADC is an acute angle, It is not possible to draw a perpendicular on it from point B.
If ADC is an obtuse angle, It is not possible to draw a perpendicular on it from point C.

My question is , How can we have both M & N simultaneously if it is not possible to draw perpendiculars to AD from both B & C simultaneously?

Am I missing something?

Experts, Please help!!!

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?

I guess I got my answer...
AD is INDEED the perpendicular to BC....This is the only possible way

So the question reduces to Is BD < CD? (a Yes/No question)

Statement 2 clearly answers the question.... Sufficient
Statement 1: Since AD is perpendicular to BC,
If AB>AC, It clearly means BD>CD
Sufficient

Answer is D

What is the OA?

nisagl750 wrote: I guess I got my answer...
AD is INDEED the perpendicular to BC....This is the only possible way

Think again. It indeed is possible even though AD were not a perpendicular to BC.

The Iceman wrote:
Think again. It indeed is possible even though AD were not a perpendicular to BC.

How..?

My reasoning is...

Let AD is not Perpendicular to BC. then D will make an angle other than 90 degree..
So Either one of the angle ADC or angle ADB will be acute and the other will be obtuse.
Let BE & CF be the perpendicular to AD, BE=M, CF=N (according to question)

Case1:
Angle ADC is acute.
In this case Angle ADB will be obtuse....
In triangle BDE
Angle EDB is obtuse (E is on AD)
Angle BED is 90 degree....
Which is not possible (sum exceeds 180 and a triangle can have at most one right angle)

Case2:
Angle ADB is acute
In this case Angle ADC will be obtuse....
Same explanation as above for tiangle CDF

What am I missing?

nisagl750 wrote: How..?

What am I missing?

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.

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.

Thats called thinking outside the Triangle.....

So, what is the OA?

nisagl750 wrote:

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.

Thats called thinking outside the Triangle.....

So, what is the OA?

OA is B.

Can some one please explain how the OA is B?

prat_agl wrote:Can some one please explain how the OA is B?

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 answers

Statement I doesn't tell us anything

=> choice (B) is the right answer