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.