# Proof: Confusing geometry problem

• Apr 6th 2008, 08:20 AM
Dslycixdued
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
• Apr 6th 2008, 08:22 AM
Mathstud28
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?
• Apr 6th 2008, 08:28 AM
Moo
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...
• Apr 6th 2008, 08:30 AM
Mathstud28
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
• Apr 6th 2008, 08:30 AM
Moo
I can understand that, i'm too often like this too :D

Now, for the problem, i admit it seems tricky oO
• Apr 6th 2008, 08:33 AM
Mathstud28
I think
this might be a tric question...because if you draw it out it seems like an impossible scenario
• Apr 6th 2008, 08:36 AM
Dslycixdued
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
• Apr 6th 2008, 08:37 AM
Moo
Well, i've got something approaching : AC is not exactly equal to BD, but it's possible
• Apr 6th 2008, 04:38 PM
Dslycixdued
Yea even when I assume AC = BD or <A = <B I just don't know what I can prove.