Thread: Proof: Confusing geometry problem

I really have no idea how to prove this

In quadrialteral ABCD the point of intersection of the perpendicular bisectors of AD and BC lies on AB. Prove that AC = BD if and only if
if you didn't realize I meant to have angle symbols but i didn't know how to make them

prove that the quadrilateral is a parallelogram...so that the diagonals which are already perpendicular will make a square proving that the diagonals are congruent...I think thats what it wants?

Hello,

If ABCD is a parallelogram, how is it possible to have the intersection on AB ? Since AD and BC are opposit, they may be parallel if it's a parallelogram. If they're parallel, their perpendicular bisectors will be parallel, and the point of intersection won't exist...

once again...for the like fourth time...I didnt read the question...I'm sorry I will be more studious in the future...I can do these things...I am just a little...uh...uncareful =D

I can understand that, i'm too often like this too

Now, for the problem, i admit it seems tricky oO

this might be a tric question...because if you draw it out it seems like an impossible scenario

Yea well the idea is to prove that its an isosceles trapezoid which is the only way AC = BD and <A has to equal <B

but I have no idea how to go about that

Well, i've got something approaching : AC is not exactly equal to BD, but it's possible

Yea even when I assume AC = BD or <A = <B I just don't know what I can prove.