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?
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
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...