Answer to Question #97267 in Analytic Geometry for Aravind

Question

Prove that in any parallelogram,the sum of the squares of all sides is equal to the sum of the squares of the diagonals

Answer 1

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.