He wants to prove that ABCD is a square with sides AB BC CD DA all equal.

Info. 1 tells us that the diagonals intersect and they have 90 degrees together, which doesn't really help.

2 tells us that the addition of the 2 opposite sides are equal. if we assume that AB+CD=BC+AD= 6 AB & CD, BC & AD could have different combinations like (4,2) (3,3) (2,4) (5,1) ..etc so we can't prove that they're all equal.

Thus there isn't sufficient info to prove that this is a square.

Koala wrote:If ABCD is quadrilateral, is AB=BC=CD=DA?

1) AC is perpendicular to BD
2) AB+CD=BC+AD

OA is E

1. AC is perpendicular to BD

Think about the geometrical shapes (as quadrilateral) that has perpendicularly intersecting diagonals.

Being a quadrilateral a square, or a Rhombus, or a Kite can have perpendicular. Though square and rhombus has all the sides equal, Kite does not have equal sides.

Thus insufficient.

2. AB+CD=BC+AD

Take simple numbers to verify.

3+4 = 2+5 (=7) so does this mean that 3 = 4 = 2 = 5? Never.

Thus insufficient.

IMO E

PS: If you have any kind of concerns about geometrical figure have a look at the below image for an idea. (Honestly I did not know what is a Kite (in terms of geometrical shape ))