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E is the answer. Let's draw B close to A (AB is relatively small, say equal to 1) and C is far away from both A and B, for example AC = 10. Then we can draw a line through B and O, the middle of AC. If D belongs to that line than BD bisects AC. Let's draw a circle with the center in C and radius AB, it intersects line BO in two points. One of those points (D') gives parallel lines R1 and R2 because angle BCA = CAD', the other one doesn't because BCA is not equal to CAD''.

I believe its a combination of 1) and 2) that makes the figure a prallelogram. Agreed? If yes then yes these lines should be parallel. C.

I disagree. There is no guarantee that it's going to be a parallelogram, solution is not unique.

A good thing to know here is that what is a minimum requirement for a quadrilateral to be a parallelogram? I thought having diagnolas bisecting and two of the opposite sides being equal was good enough a reason to assume this quadrilateral was a parallelogram.

A good thing to know here is that what is a minimum requirement for a quadrilateral to be a parallelogram? I thought having diagnolas bisecting and two of the opposite sides being equal was good enough a reason to assume this quadrilateral was a parallelogram.

It says BD bisects AC. It doesn't say anywhere that BD and AC bisect each other.

A good thing to know here is that what is a minimum requirement for a quadrilateral to be a parallelogram? I thought having diagnolas bisecting and two of the opposite sides being equal was good enough a reason to assume this quadrilateral was a parallelogram.

It says BD bisects AC. It doesn't say anywhere that BD and AC bisect each other.

Sure! but we have two properties here

1) AB = DC 2) BD bisects AC

Can AB be equal to DC if AC also did not bisect BD? I was not able to come up with a case...

A good thing to know here is that what is a minimum requirement for a quadrilateral to be a parallelogram? I thought having diagnolas bisecting and two of the opposite sides being equal was good enough a reason to assume this quadrilateral was a parallelogram.

It says BD bisects AC. It doesn't say anywhere that BD and AC bisect each other.

Sure! but we have two properties here

1) AB = DC 2) BD bisects AC

Can AB be equal to DC if AC also did not bisect BD? I was not able to come up with a case...

Follow my example, posted above. E is the answer. Let's draw B close to A (AB is relatively small, say equal to 1) and C is far away from both A and B, for example AC = 10. Then we can draw a line through B and O, the middle of AC. If D belongs to that line than BD bisects AC. Let's draw a circle with the center in C and radius AB, it intersects line BO in two points. One of those points (D') gives parallel lines R1 and R2 because angle BCA = CAD', the other one doesn't because BCA is not equal to CAD''.

Statement 2, by itself, tells us that AB must equal BC and that AD must equal CD. Imagining a kite (in which the vertical axis bisects the horizontal axis (but not vice-versa)) may make the rule easier to understand. Combining this conclusion with statement 1 indicates that all four segments in the perimeter must be equal (transitive rule of equalities), therefore, the figure is a rhombus (but not necessarily a square). Rhombuses are a class of parallelogram, so AC then becomes a transversal of two parallel lines so angle BCA = angle CAD.

Answer: C.

gmatclubot

Re: Geometry Problem
[#permalink]
23 Jul 2008, 04:40