urgent!!!plz answer as soon as possible

 

AEB and AED are right anlged triangles, with AEB = AED = 90         ....Diagonals intersect at right angles.

AE² + EB² =  AB² 

AE² + ED² =  AD²       ...........Pythagorus's theorem.

Similarly for  CEB and CED

CE² + EB² =  BC² 

CE² + ED² =  CD²       ...........Pythagorus's theorem.

Adding all four equation,

AE² + EB² + AE² + ED² + CE² + EB² + CE² + ED² = AB²+BC²+CD²+AD² ...........(1)

Now,

AC² = (AE+EC)² = AE²+EC²+2AEEC = AE²+EC²+2AE²  ....................Diagonals bisect each other.

= AE² + EC²+ AE² +EC²         ........writing out 2AE² = AE² + AE² = AE² + EC²

Similarly,

BD² = BE² + ED² + BE² + ED²

Using this result in (1).

AC² + BD² = AB²+BC²+CD²+AD²

Regards,

Team,

TopperLearning.