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by voodoo_child » Mon May 23, 2011 10:22 am
If A, B, C and D form a quadrilateral.

Is AC > BD ?

a) Angle ABC < Angle BCD
b) AB = BC= CD = DA

I was to able to figure out that b) is insufficient. Any thoughts about a) ?

OA - c

thanks
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by djiddish98 » Mon May 23, 2011 11:39 am
I can't think of any geometry proof about angles, but try drawing a trapezoid with bases AB and DC.

If you make the line DC very short with and close to point B, you'll have a longer AC diagonal than DB.

If you shift DC towards A, you'll have a shorter AC diagonal than DB.

I think the point is that the length of a diagonal is going to depend on both the angle as well as the length of the other two sides.
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by GMATGuruNY » Mon May 23, 2011 12:07 pm
voodoo_child wrote:If A, B, C and D form a quadrilateral.

Is AC > BD ?

a) Angle ABC < Angle BCD
b) AB = BC= CD = DA

I was to able to figure out that b) is insufficient. Any thoughts about a) ?

OA - c

thanks
Voodoo
Statement 1: ∠ABC < ∠BCD.

∠ABC < ∠BCD and AC<BD:

Image

∠ABC < ∠BCD and AC>BD:

Image
Insufficient.

Statement 2: AB=BC=CD=DA.
If ABCD is a square, then AC=BD.
If ABCD is a rhombus that is not a square, then one diagonal is longer than the other. No way to determine whether AC>BD or AC<BD.
Insufficient.

Statement 1 and 2 together: ∠ABC < ∠BCD and AB=BC=CD=DA.
Since all the sides are equal and ∠ABC < ∠BCD, we know that AC<BD:

Image

The correct answer is C.
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by cans » Wed May 25, 2011 8:19 pm
b) ab=bc=cd=da it can be rhombus (diagonals not equal) and a square (diagonals equal)
so b alone is not sufficient..
a) angle abc < angle bcd
this also is not sufficient.
But if we combine both of them, all sides equal and angle abc < angle bcd,
it means we have a rhombus and as angle abc < angle bcd,
we have ac<bd..... this both together are sufficient.
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