# Thread: Proof: Confusing geometry problem

1. ## 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 $\displaystyle <A = <B$
if you didn't realize I meant to have angle symbols but i didn't know how to make them

2. ## Your jsut tryign to

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?

3. 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...

4. ## Yeah

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

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

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

6. ## I think

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

7. ## Its a trapezoid (Im pretty sure)

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

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

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