Louis14
A diagonal of a parallelogram divides the parallelogram into two congruent triangles, but does it also mean that that diagonal also bisects the angles through which it originates? For example, we know that the diagonal of a rectangle would divide the rectangle into two congruent (equal) triangles, but, at the same time, would that diagonal also be bisecting the angles through which it originates? In other words, can a parallelogram have a diagonal that creates two congruent triangles but doesn't bisect the angles through which it originates?
In general, a good approach for concerns like this is to draw some pictures and test it out. I encourage you to do so even though you've already had your question answered, just because this is the same kind of "case testing" that is incredibly useful while actually doing GMAT geometry problems. Specifically, can you draw an "extreme" rectangle (or parallelogram) where it's completely clear that the diagonals
don't bisect the angles? Or can you draw one where the converse is true: it's totally obvious (or even mathematically provable) that the diagonals
do bisect the angles?
For the first, try drawing a very long, skinny rectangle, then sketch its diagonals. The diagonals should cut each 90 degree corner of the rectangle into one large angle, and one much smaller angle. For the second, if you draw a square, you can use what you know about right triangles to show that the diagonals cut each 90 degree corner into two identical 45 degree angles.