Question

The diagonals of quadrilateral ABCD intersect at point K. Is each of the following needed to prove that ABCD is a parallelogram?

Answer 1

BK=AK: no

BK=DK:yes

CK=AK:yes

CK=DK:no

Explanation:

for the first and last, it is not needed because a rectangle is a parallelogram, for example, but its diagonals are not congruent. The diagonals intersect each other at both of their midpoints, on the other hand, so therefore the second and third must be true. Does this make sense?