Answer to Question #97267 in Analytic Geometry for Aravind

Question #97267
Prove that in any parallelogram,the sum of the squares of all sides is equal to the sum of the squares of the diagonals
1
Expert's answer
2019-10-25T11:29:24-0400

First we draw the perpendiculars from C and D on AB as can be seen in the attached figure.

In the right triangle AEC we apply Pythagorean Theorem: AC2=AE2+CE2 which implies

AC2=(AB+BE)2 +CE2 = AB2 + 2*AB*BE+BE2 +CE2 (*)

Since CDFE is a rectangle, it follows that CD=EF.

But AB=CD because ABCD is parallelogram. So it follows that AB=CD=EF.

But DF=CE being heights in ABCD and represent the distance between parallels AB and CD.

Also AD=BC since ABCD is parallelogram.

Thus the right triangles AFD and BCE are congruent. So AF=BE.

Now in triangle DFB: BD2 = BF2 +FD2 by Pythagorean Theorem.

So BD2 = (AB-AF)2 + CE2 = (AB-BE)2 +CE2 = AB2 -2*AB*BE+BE2 + CE2 (**)


Adding relationships (*) and (**): AC2+BD2 = 2AB2 +2BE2 +2CE2 =2AB2 +2(BE2+CE2) =2AB2 +2BC2 after using

Pythagorean Theorem in triangle BCE.

In conclusion, AC2+BD2 = AB2 +BC2+CD2+DA2 as required.

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