urgent!!!plz answer as soon as possible

Asked by honey_physics | 14th Feb, 2010, 11:35: AM

Expert Answer:

 

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.

Answered by  | 14th Feb, 2010, 12:31: PM

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